Projectiles - derived equations

1. Projectile model: acceleration components and notation

FORMULA

$$a_x=0$$

$$a_y=-g$$

DERIVATION

1. Only weight acts during free flight.

2. Use Newton's second law horizontally.

$$F_x=ma_x=0\quad\Rightarrow\quad a_x=0$$

3. Take upwards as positive.

$$F_y=ma_y=-mg\quad\Rightarrow\quad a_y=-g$$

ASSUMPTIONS

No air resistance. Constant g. Launch is at (0, 0). Angled launches use 0 < θ < 90° unless stated otherwise.

SYMBOLS

u = launch speed

θ = launch angle above the horizontal

t = time since launch

x = horizontal displacement

y = vertical displacement above launch

g = positive gravitational acceleration

H = maximum rise above launch

T = total flight time

R = horizontal range

h = positive drop to the landing level

2. Initial horizontal and vertical velocity components

FORMULA

$$u_x=u\cos\theta$$

$$u_y=u\sin\theta$$

DERIVATION

1. Draw a right-angled velocity triangle. The hypotenuse is the launch speed u.

2. Use the cosine ratio for the horizontal component.

$$\cos\theta=\frac{u_x}{u}$$

$$u_x=u\cos\theta$$

3. Use the sine ratio for the vertical component.

$$\sin\theta=\frac{u_y}{u}$$

$$u_y=u\sin\theta$$

CHECK

The component squares add to the square of the launch speed.

3. Horizontal velocity at any time

FORMULA

$$v_x=u\cos\theta$$

DERIVATION

1. Use the velocity equation horizontally.

$$v_x=u_x+a_xt$$

2. Substitute the initial component and zero horizontal acceleration.

$$v_x=u\cos\theta+0t$$

3. Simplify.

$$v_x=u\cos\theta$$

WHAT IT MEANS

Horizontal velocity stays constant throughout free flight.

4. Vertical velocity at any time

FORMULA

$$v_y=u\sin\theta-gt$$

DERIVATION

1. Use the velocity equation vertically.

$$v_y=u_y+a_yt$$

2. The initial vertical component is u sinθ. Upwards is positive.

$$u_y=u\sin\theta\quad,\quad a_y=-g$$

3. Substitute.

$$v_y=u\sin\theta-gt$$

WHAT IT MEANS

Vertical velocity is positive on ascent, zero at the peak and negative on descent.

5. Horizontal displacement at any time

FORMULA

$$x=(u\cos\theta)t$$

DERIVATION

1. Use the displacement equation horizontally.

$$x=u_xt+\frac{1}{2}a_xt^2$$

2. Substitute the initial component and zero acceleration.

$$x=(u\cos\theta)t+\frac{1}{2}(0)t^2$$

3. Simplify. You can then make time the subject.

$$x=(u\cos\theta)t$$

$$t=\frac{x}{u\cos\theta}$$

REMEMBER

Use this same time in the vertical equations. The rearrangement requires a nonzero horizontal velocity.

6. Vertical displacement at any time

FORMULA

$$y=(u\sin\theta)t-\frac{1}{2}gt^2$$

DERIVATION

1. Use the displacement equation vertically.

$$y=u_yt+\frac{1}{2}a_yt^2$$

2. Substitute the initial vertical component and acceleration.

$$u_y=u\sin\theta\quad,\quad a_y=-g$$

3. Simplify.

$$y=(u\sin\theta)t-\frac{1}{2}gt^2$$

SIGN CHECK

y is measured from launch. A landing h below launch has y = −h. If launch is h above ground, height above ground is h + y.

7. Vertical velocity and vertical displacement, without time

FORMULA

$$v_y^2=u^2\sin^2\theta-2gy$$

DERIVATION

1. Use the SUVAT equation without time, vertically.

$$v_y^2=u_y^2+2a_y y$$

2. Substitute the initial component and downward acceleration.

$$v_y^2=(u\sin\theta)^2+2(-g)y$$

3. Simplify.

$$v_y^2=u^2\sin^2\theta-2gy$$

REMEMBER

This gives the vertical component, not the total speed. Squaring loses the direction, so choose rising or falling separately.

8. Projectile trajectory: vertical displacement in terms of horizontal displacement

FORMULA

$$y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$$

DERIVATION

1. Make time the subject of the horizontal equation.

$$t=\frac{x}{u\cos\theta}$$

2. Substitute that time into the vertical equation.

$$y=(u\sin\theta)\left(\frac{x}{u\cos\theta}\right)-\frac{g}{2}\left(\frac{x}{u\cos\theta}\right)^2$$

3. Cancel u in the first term and square the fraction in the second.

$$y=\frac{x\sin\theta}{\cos\theta}-\frac{gx^2}{2u^2\cos^2\theta}$$

4. Use the tangent identity.

$$\frac{\sin\theta}{\cos\theta}=\tan\theta$$

WHAT IT MEANS

The path is a downward-opening parabola. The equation requires nonzero horizontal velocity.

FRACTION CHECK

Only the second term is subtracted as a fraction. Its whole numerator is gx² and its whole denominator is 2u² cos²θ.

9. Time to reach maximum height

FORMULA

$$t_{\mathrm{peak}}=\frac{u\sin\theta}{g}$$

DERIVATION

1. At maximum height, vertical velocity is zero.

$$v_y=0$$

2. Use the vertical velocity equation at that time.

$$0=u\sin\theta-gt_{\mathrm{peak}}$$

3. Rearrange.

$$gt_{\mathrm{peak}}=u\sin\theta$$

$$t_{\mathrm{peak}}=\frac{u\sin\theta}{g}$$

WHEN TO USE IT

An upward launch that reaches its peak before a collision.

REMEMBER

Vertical velocity is zero at the peak. Acceleration is still −g.

10. Maximum height above the launch point

FORMULA

$$H=\frac{u^2\sin^2\theta}{2g}$$

DERIVATION

1. At the highest point, set vertical velocity to zero and displacement to H.

$$v_y=0\quad,\quad y=H$$

2. Use the vertical SUVAT equation.

$$v_y^2=u_y^2+2a_y y$$

3. Substitute the components and acceleration.

$$0=(u\sin\theta)^2-2gH$$

4. Rearrange.

$$2gH=u^2\sin^2\theta$$

$$H=\frac{u^2\sin^2\theta}{2g}$$

WHEN TO USE IT

An upward launch that reaches its peak before a collision.

REMEMBER

H is the rise above launch. Add the launch height to get maximum height above ground.

11. Horizontal distance to the highest point

FORMULA

$$x_{\mathrm{peak}}=\frac{u^2\sin\theta\cos\theta}{g}$$

$$x_{\mathrm{peak}}=\frac{u^2\sin(2\theta)}{2g}$$

DERIVATION

1. Use the time to the highest point.

$$t_{\mathrm{peak}}=\frac{u\sin\theta}{g}$$

2. Substitute into the horizontal displacement equation.

$$x_{\mathrm{peak}}=(u\cos\theta)\left(\frac{u\sin\theta}{g}\right)$$

3. Simplify, then use the double-angle identity.

$$x_{\mathrm{peak}}=\frac{u^2\sin\theta\cos\theta}{g}$$

$$2\sin\theta\cos\theta=\sin(2\theta)$$

WHEN TO USE IT

The projectile must reach its peak before a collision.

SAME-HEIGHT LANDING

The highest point is halfway across the range: x at the peak = R/2.

12. Minimum speed during a flight that reaches the peak

FORMULA

$$v_{\min}=u\cos\theta$$

DERIVATION

1. Combine the perpendicular velocity components.

$$v^2=v_x^2+v_y^2$$

2. The horizontal component is constant, so speed is smallest when the vertical component is zero.

$$v_y=0\quad\Rightarrow\quad v_{\min}^2=v_x^2$$

3. At the peak, substitute the horizontal component.

$$v_{\min}=\sqrt{u^2\cos^2\theta}=u\cos\theta$$

WHEN TO USE IT

An upward launch with positive horizontal velocity that reaches its peak.

REMEMBER

Only the vertical velocity is zero there. The projectile is still moving horizontally.

13. Total flight time when launch and landing are at the same height

FORMULA

$$T=\frac{2u\sin\theta}{g}$$

DERIVATION

1. Landing at launch height means zero vertical displacement.

$$y=0$$

2. Use the vertical displacement equation.

$$0=(u\sin\theta)T-\frac{1}{2}gT^2$$

3. Factorise.

$$0=T\left(u\sin\theta-\frac{1}{2}gT\right)$$

4. Ignore T = 0, which is launch. Solve the other factor.

$$u\sin\theta-\frac{1}{2}gT=0$$

$$T=\frac{2u\sin\theta}{g}$$

WHEN TO USE IT

Launch and landing must be at the same height.

REMEMBER

The total flight time is twice the time to the peak. Ascent and descent take equal times.

14. Sine double-angle identity used in the range equation

FORMULA

$$\sin(2\theta)=2\sin\theta\cos\theta$$

DERIVATION

1. Start from the sine addition identity.

$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$

2. Set both angles equal to θ.

$$\sin(\theta+\theta)=\sin\theta\cos\theta+\cos\theta\sin\theta$$

3. Combine the two identical terms.

$$\sin(2\theta)=2\sin\theta\cos\theta$$

WHY IT IS USEFUL

It makes the horizontal range equation shorter.

15. Horizontal range when launch and landing are at the same height

FORMULA

$$R=\frac{u^2\sin(2\theta)}{g}$$

DERIVATION

1. Horizontal range is horizontal velocity multiplied by flight time.

$$R=(u\cos\theta)T$$

2. For landing at launch height, use the total flight time.

$$T=\frac{2u\sin\theta}{g}$$

3. Substitute and simplify.

$$R=(u\cos\theta)\left(\frac{2u\sin\theta}{g}\right)$$

$$R=\frac{2u^2\sin\theta\cos\theta}{g}$$

4. Use the double-angle identity.

$$R=\frac{u^2\sin(2\theta)}{g}$$

WHEN TO USE IT

Launch and landing must be at the same height.

REMEMBER

R is horizontal distance. It is not the length of the curved flight path.

16. Launch angle for maximum horizontal range on level ground

FORMULA

$$\theta=45^\circ$$

$$R_{\max}=\frac{u^2}{g}$$

DERIVATION

1. Start with the range for a same-height landing.

$$R=\frac{u^2}{g}\sin(2\theta)$$

2. With u and g fixed, range is greatest when the sine factor is 1.

$$\sin(2\theta)=1$$

3. Choose the maximum within the allowed launch angles.

$$2\theta=90^\circ\quad\Rightarrow\quad\theta=45^\circ$$

4. Substitute the maximum sine value.

$$R_{\max}=\frac{u^2}{g}$$

WHEN TO USE IT

Fixed launch speed and equal launch and landing heights. For unequal heights, 45° is generally not the best angle.

17. Complementary launch angles give the same range on level ground

FORMULA

$$R(\theta)=R(90^\circ-\theta)$$

DERIVATION

1. Write the range for the first angle.

$$R(\theta)=\frac{u^2}{g}\sin(2\theta)$$

2. Write it for the complementary angle.

$$R(90^\circ-\theta)=\frac{u^2}{g}\sin(180^\circ-2\theta)$$

3. Use the sine symmetry identity.

$$\sin(180^\circ-A)=\sin A$$

4. The ranges are therefore equal.

$$R(90^\circ-\theta)=\frac{u^2}{g}\sin(2\theta)=R(\theta)$$

WHEN TO USE IT

Same launch speed, same g and both land at launch height.

REMEMBER

The larger launch angle gives a greater maximum height and a longer flight time.

18. Maximum height in terms of total flight time on level ground

FORMULA

$$H=\frac{gT^2}{8}$$

DERIVATION

1. Rearrange the same-height flight-time equation.

$$T=\frac{2u\sin\theta}{g}\quad\Rightarrow\quad u\sin\theta=\frac{gT}{2}$$

2. Substitute into the maximum-height equation.

$$H=\frac{(u\sin\theta)^2}{2g}$$

$$H=\frac{\left(\frac{gT}{2}\right)^2}{2g}$$

3. Square the whole numerator and simplify.

$$H=\frac{g^2T^2}{8g}=\frac{gT^2}{8}$$

WHEN TO USE IT

T is the whole flight time back to launch height, not just the time to the peak.

19. Maximum height in terms of range and launch angle on level ground

FORMULA

$$H=\frac{R\tan\theta}{4}$$

DERIVATION

1. Write the maximum-height and same-height range equations.

$$H=\frac{u^2\sin^2\theta}{2g}$$

$$R=\frac{2u^2\sin\theta\cos\theta}{g}$$

2. Divide the height by the range. Dividing by a fraction means multiplying by its reciprocal.

$$\frac{H}{R}=\frac{u^2\sin^2\theta}{2g}\times\frac{g}{2u^2\sin\theta\cos\theta}$$

3. Cancel common factors and use the tangent identity.

$$\frac{H}{R}=\frac{\sin\theta}{4\cos\theta}=\frac{\tan\theta}{4}$$

4. Multiply by R.

$$H=\frac{R\tan\theta}{4}$$

WHEN TO USE IT

Equal launch and landing heights, with 0 < θ < 90°.

20. Total speed at any time

FORMULA

$$v=\sqrt{(u\cos\theta)^2+(u\sin\theta-gt)^2}$$

DERIVATION

1. Use Pythagoras for the perpendicular velocity components.

$$v^2=v_x^2+v_y^2$$

2. Substitute the two components.

$$v^2=(u\cos\theta)^2+(u\sin\theta-gt)^2$$

3. Take the positive square root because speed is a magnitude.

$$v=\sqrt{(u\cos\theta)^2+(u\sin\theta-gt)^2}$$

OPTIONAL EXPANSION

The same result can be written as v² = u² − 2ugt sinθ + g²t².

21. Total speed at a given vertical displacement

FORMULA

$$v^2=u^2-2gy$$

$$v=\sqrt{u^2-2gy}$$

DERIVATION

1. Write the squared vertical component.

$$v_y^2=u^2\sin^2\theta-2gy$$

2. Write the squared horizontal component.

$$v_x^2=u^2\cos^2\theta$$

3. Add the component squares.

$$v^2=u^2(\cos^2\theta+\sin^2\theta)-2gy$$

4. Use the trigonometric identity.

$$\cos^2\theta+\sin^2\theta=1$$

$$v^2=u^2-2gy$$

CONSEQUENCES

Speed is u again at launch height. At a drop h below launch, y = −h and v² = u² + 2gh. The projectile must actually reach the chosen height.

22. Direction of the velocity at any time

FORMULA

$$\tan\phi=\frac{u\sin\theta-gt}{u\cos\theta}$$

$$\phi=\tan^{-1}\left(\frac{u\sin\theta-gt}{u\cos\theta}\right)$$

DERIVATION

1. Draw the instantaneous velocity triangle. Measure φ from the positive horizontal.

$$\tan\phi=\frac{v_y}{v_x}$$

2. Substitute the components.

$$\tan\phi=\frac{u\sin\theta-gt}{u\cos\theta}$$

3. Take inverse tangent.

$$\phi=\tan^{-1}\left(\frac{u\sin\theta-gt}{u\cos\theta}\right)$$

WHEN TO USE IT

The horizontal velocity is positive, as in this set.

REMEMBER

Positive φ means rising. Zero means the peak. Negative φ means falling.

23. Time or times at which a projectile reaches a specified height

FORMULA

$$t=\frac{u\sin\theta\pm\sqrt{u^2\sin^2\theta-2gY}}{g}$$

DERIVATION

1. Y is the required signed vertical displacement from launch.

$$Y=(u\sin\theta)t-\frac{1}{2}gt^2$$

2. Rearrange into a quadratic.

$$\frac{1}{2}gt^2-(u\sin\theta)t+Y=0$$

3. Use the quadratic formula.

$$t=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$$

$$A=\frac{g}{2}\quad,\quad B=-u\sin\theta\quad,\quad C=Y$$

4. Substitute and simplify.

$$t=\frac{u\sin\theta\pm\sqrt{u^2\sin^2\theta-2gY}}{g}$$

INTERPRETING THE ANSWERS

For a height between launch and the peak, the minus root is on ascent and the plus root is on descent. At the peak they coincide. A negative discriminant means the height cannot be reached. Keep only nonnegative times before a collision.

24. Total flight time when landing a vertical distance h below launch

FORMULA

$$T=\frac{u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}}{g}$$

DERIVATION

1. h is a positive drop, so the signed vertical displacement is −h.

$$-h=(u\sin\theta)T-\frac{1}{2}gT^2$$

2. Rearrange.

$$\frac{1}{2}gT^2-(u\sin\theta)T-h=0$$

3. Use the quadratic formula.

$$T=\frac{u\sin\theta\pm\sqrt{u^2\sin^2\theta+2gh}}{g}$$

4. The minus root is negative. Choose the plus root.

$$T=\frac{u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}}{g}$$

CHECK

For an upward launch, setting h = 0 recovers the same-height flight time.

FRACTION CHECK

The entire sum, including the square root, is divided by g.

25. Horizontal range when landing a vertical distance h below launch

FORMULA

$$R=\frac{u\cos\theta}{g}\left(u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}\right)$$

DERIVATION

1. Horizontal range is horizontal velocity multiplied by flight time.

$$R=(u\cos\theta)T$$

2. Use the flight time for landing h below launch.

$$T=\frac{u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}}{g}$$

3. Substitute the whole expression for time.

$$R=(u\cos\theta)\left(\frac{u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}}{g}\right)$$

4. Equivalently, put the entire product over g.

$$R=\frac{u\cos\theta\left(u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}\right)}{g}$$

WHEN TO USE IT

h is a positive drop to a horizontal landing level.

CHECK

For an upward launch, h = 0 gives the same-height range equation.

26. Time of flight for a horizontal launch from height h

FORMULA

$$T=\sqrt{\frac{2h}{g}}$$

DERIVATION

1. A horizontal launch has zero initial vertical velocity. Take downwards as positive for this derivation.

$$u_y=0\quad,\quad a_y=g$$

2. Use vertical displacement.

$$h=0T+\frac{1}{2}gT^2$$

3. Rearrange.

$$T^2=\frac{2h}{g}$$

4. Choose the positive time.

$$T=\sqrt{\frac{2h}{g}}$$

WHEN TO USE IT

Horizontal launch, landing h below launch.

SQUARE-ROOT CHECK

Both 2h and g are inside the square root.

WHAT IT MEANS

For a fixed h, flight time does not depend on the horizontal launch speed.

27. Horizontal range for a horizontal launch from height h

FORMULA

$$R=u\sqrt{\frac{2h}{g}}$$

DERIVATION

1. For a horizontal launch, all the initial speed is horizontal.

$$v_x=u$$

2. Get the flight time from the vertical motion.

$$T=\sqrt{\frac{2h}{g}}$$

3. Use horizontal distance = horizontal speed × time.

$$R=uT$$

4. Substitute the flight time.

$$R=u\sqrt{\frac{2h}{g}}$$

WHEN TO USE IT

Horizontal launch, landing h below launch.

SQUARE-ROOT CHECK

The launch speed u is outside the square root. The whole fraction 2h/g is inside it.

28. Trajectory equation for a horizontal launch

FORMULA

$$y=-\frac{gx^2}{2u^2}$$

DERIVATION

1. Horizontal motion gives the time.

$$x=ut\quad\Rightarrow\quad t=\frac{x}{u}$$

2. The initial vertical velocity is zero. Use upwards as positive.

$$y=-\frac{1}{2}gt^2$$

3. Substitute the time and square the whole fraction.

$$y=-\frac{g}{2}\left(\frac{x}{u}\right)^2$$

$$y=-\frac{gx^2}{2u^2}$$

SIGN CHECK

y is measured upwards from launch, so it is negative below launch. A positive downward drop d instead satisfies d = −y.

29. Impact speed for a horizontal launch from height h

FORMULA

$$v=\sqrt{u^2+2gh}$$

DERIVATION

1. The horizontal component stays constant.

$$v_x=u$$

2. Use the vertical SUVAT equation with downwards positive.

$$v_y^2=0^2+2gh=2gh$$

3. Add the squared components.

$$v^2=v_x^2+v_y^2=u^2+2gh$$

4. Take the positive square root.

$$v=\sqrt{u^2+2gh}$$

WHEN TO USE IT

Horizontal launch, landing h below launch.

SQUARE-ROOT CHECK

The whole sum u² + 2gh is inside the square root.

30. Vertical drop below the straight-line path the projectile would follow without gravity

FORMULA

$$d=\frac{1}{2}gt^2$$

$$d=\frac{gx^2}{2u^2\cos^2\theta}$$

DERIVATION

1. Without gravity, the height at the same time would be:

$$y_0=(u\sin\theta)t$$

2. With gravity, the actual height is:

$$y=(u\sin\theta)t-\frac{1}{2}gt^2$$

3. Subtract to find the vertical drop below the initial straight-line path.

$$d=y_0-y=\frac{1}{2}gt^2$$

4. Substitute the time from horizontal motion.

$$t=\frac{x}{u\cos\theta}$$

$$d=\frac{g}{2}\left(\frac{x}{u\cos\theta}\right)^2=\frac{gx^2}{2u^2\cos^2\theta}$$

REMEMBER

d is measured vertically below the initial straight-line direction, not perpendicular to that line. It is not generally the height lost since launch