Notes: Vertical Motion under Constant Gravity (Upward Throw at 22 m/s)

Coordinate System and Sign Convention

• We use a vertical y-axis with positive upwards.
• The acceleration for all free-fall problems is -g (downward).
• We can rewrite the three equations we had earlier by replacing x with y and a with -g, giving new equivalent equations for vertical motion with known acceleration.
• Because the acceleration is known, there are only five unknown variables in the setup.
• The velocity of a ball thrown upwards is positive on the way up and negative on the way down.
• At the turning point (apex), the velocity is zero.
• Gravity on a planet doesn’t flip direction or turn off during the flight; it remains downward with the same magnitude g (within the same planet).
• The problem statement: a ball is thrown straight up into the air at 22 m/s. How high does it go? We will work through the setup and solution.

Kinematic relationships for vertical motion

• Basic definitions (derivatives):
  • $v = \frac{dy}{dt}, \quad a = \frac{dv}{dt} = \frac{d^2 y}{dt^2}$
• With constant downward acceleration a = -g:
  • $v(t) = v_0 - g t$
  • $y(t) = y<em>0 + v</em>0 t - \frac{1}{2} g t^2$
  • $a = -g$
• Key qualitative points:
  • Acceleration is constant and downward for all times during the motion.
  • The velocity changes sign as the ball rises and then falls; the apex occurs when v = 0.

Problem setup: ball thrown upward at 22 m/s

• Given initial velocity: $v_0 = 22\ \mathrm{m}/\mathrm{s}$
• Initial height: $y_0 = 0\ \mathrm{m}$ (launch height)
• Question: How high does it go (maximum height, $h_{\max}$)?
• Relevant relation tying v and y for constant a (useful form):
  • $v^2 = v<em>0^2 + 2 a (y - y</em>0)$
  • With apex v = 0 and a = -g, let (\Delta y = h_{\max}):
  • $0 = v<em>0^2 + 2(-g) h</em>{\max}$
  • $h<em>{\max} = \frac{v</em>0^2}{2 g}$

Calculation of maximum height for v0 = 22 m/s

• Substitute values:
  • $h_{\max} = \frac{(22)^2}{2 g} = \frac{484}{2 g} = \frac{242}{g} \ \text{m}$
• With standard Earth gravity $g \approx 9.8 \ \mathrm{m}/\mathrm{s}^2$:
  • $h_{\max} \approx \frac{242}{9.8} \approx 24.7 \ \mathrm{m}$
• Alternative exact expression (explicit numeric using g):
  • $h<em>{\max} = \frac{v</em>0^2}{2 g} = \frac{22^2}{2 \cdot 9.8} \ \text{m} \approx 24.7 \ \text{m}$

Time to reach apex and velocity-time relationship

• Time to reach the apex:
  • $t<em>{\text{apex}} = \frac{v</em>0}{g} = \frac{22}{9.8} \approx 2.24\ \text{s}$
• The velocity as a function of time:
  • $v(t) = v_0 - g t = 22 - 9.8 t$
• The height as a function of time:
  • $y(t) = y<em>0 + v</em>0 t - \frac{1}{2} g t^2 = 0 + 22 t - \frac{1}{2} (9.8) t^2$

Additional related results (optional extensions)

• Total flight time (neglecting air resistance):
  • $t<em>{\text{total}} = 2 t</em>{\text{apex}} = \frac{2 v_0}{g} = \frac{2 \cdot 22}{9.8} \approx 4.49\ \text{s}$
• Maximum height formula intuition: height increases with the square of the initial speed and inversely with gravity.
• For other planets, replace g with the planet’s gravitational acceleration; the same kinematic framework applies.

Key takeaways and practical implications

• In vertical motion with constant gravity, the motion can be fully described by a small set of equations:
  • $v(t) = v_0 - g t$
  • $y(t) = y<em>0 + v</em>0 t - \frac{1}{2} g t^2$
  • $a = -g$
• The apex occurs when $v = 0$, leading to the simple expression for maximum height $h<em>{\max} = \dfrac{v</em>0^2}{2 g}$.
• The sign convention (positive up, velocity positive up, acceleration negative) ensures consistent results across time as the ball moves upward and then downward.
• This framework connects directly to Newton’s laws under a constant gravitational field and provides a foundation for more complex motion (e.g., including air resistance or horizontal components).