Projectile Motion

what is a projectile

anything that is thrown

• if it is thrown downwards, it will feel the Earth’s pull due to gravity (acceleration in the y direction is -g)

motion of a projectile

if air resistance is neglected, then the motion of a projectile is the superposition of these independent motions:

• vx: a horizontal motion with constant velocity, which is the initial horizontal velocity v0x = v0 cos θ

• vy: a vertical motion with constant acceleration ay = -g and initial vertical velocity v0y = v0 sin θ, thus vy = v0 sin θ - gt

equations of motion of a projectile

• xt = x0 + (v0 cos θ0)t

• yt = y0 + (v0 sin θ0)t - ½ gt²

  • x0 and y0 are the initial coordinates

  • vector v0 is the initial velocity (v0 cos θ + v0 sin θ)

  • θ0 is the initial angle (aim)

from the equation v0 sin θ - gt, we can determine

at what time will v0y = 0 m/s (the projectile is at its maximum height)

• find by setting vy = 0, thus 0 = v0 sin θ - gt

the moment an object is thrown as a projectile, it has a

constant downward acceleration due to gravity (-g)

• the horizontal velocity is constant (a = 0), and vertical velocity is changing due to acceleration being constant (a = -9.80 m/s²)

time of flight is represented by

tf, which is the total time that the projectile spends in the air (neglecting air resistance)

• depends only on the vertical component of the initial velocity, v0y

when the projectile hits the ground,

y = 0, therefore y(tf) = (v0 sin θ)tf - ½ gtf² = 0

• thus, tf = 2(v0 sin θ) / g (vertical velocity component v0y)

horizontal range of motion

range from the initial to final x position

• the horizontal range is R which equals x(tf), so R = x(tf) = (v0 cos θ)tf

  • plugging in the tf value gives (v0 cos θ) x (2(v0 sin θ) / g)

• thus, R = 2(v0² sin θ cos θ) / g

  • since sin 2 θ = 2 sin θ cos θ, R = v0² sin 2 θ / g

  • for a given initial velocity, the range is maximized if θ = 45° (2 θ = 90°)

• R max = v0² / g

the horizontal range formula for a projectile is only valid when the

launch height is equal to the landing height

• since the time of flight derives the results, and tf is the time the projectile takes to make a complete parabola in the air (same time to reach max height from the origin as it takes to reach the origin from the max height)

for the entire parabola for projectile motion

• 0 = v0y t - ½ gt², thus t = 2 v0y / g

  • t = 2 v0 sin θ / g

at the midpoint (highest point) for projectile motion

• vector v = vector vx, since vy = 0

• tf = 2 v0y / g

• t = tf / 2, and ∆x = vx tf / 2 = 2 vx v0y / 2g, thus ∆x = vx v0y / g

  • ∆x = (v0 cos θ) x (v0 sin θ) / g = v0² sin θ cos θ / g