Free-Fall Acceleration — Question 2

Question 2

Equal impact speeds do not imply equal flight times

Two small balls of different masses are released simultaneously from the same height $H=30\ \mathrm{m}$ above the ground in separate vertical lanes. Ball U is thrown upward at $u=10\ \mathrm{m/s}$; ball D is thrown downward at the same speed. Treat each as a point in free fall with constant $g=10\ \mathrm{m/s^2}$ and no air resistance. Use ground-based upward-positive coordinates.

1. Derive each positive ground-contact time in terms of $H,u,g$, then evaluate them. Find their difference without subtracting rounded numbers.

2. Determine both velocities immediately before impact. Why can their speeds agree even though their flight times differ?

3. Explain the time difference by considering U’s return to its release height. Does mass enter any of these results?

4. A student asserts that U spends equal times rising and falling because free fall is symmetric. State the correct symmetry and identify the unequal heights that invalidate this claim for the whole flight.

Question 2 - Solution

Equal impact speeds do not imply equal flight times

(a) Keep the sign of each initial velocity.

The trajectories are $y_U=H+ut-gt^2/2$ and $y_D=H-ut-gt^2/2$. Their positive roots are $$\boxed{T_U=\frac{u+\sqrt{u^2+2gH}}{g}},\qquad \boxed{T_D=\frac{-u+\sqrt{u^2+2gH}}{g}}.$$ Here $\sqrt{u^2+2gH}=10\sqrt7\ \mathrm{m/s}$, so $$T_U=1+\sqrt7\approx3.646\ \mathrm{s},\quad T_D=-1+\sqrt7\approx1.646\ \mathrm{s}.$$ Thus $\boxed{T_U-T_D=2u/g=2.00\ \mathrm{s}}$ exactly in the stated model.

(b) Select the downward branch at ground contact.

For either ball, $v^2=u^2+2gH$. Both are descending on contact, hence $$\boxed{v_U=v_D=-10\sqrt7\approx-26.46\ \mathrm{m/s}}.$$ The squared relation depends on initial speed and net height change, while the time equations also retain initial direction and the extra upward excursion.

(c) Match states at the same height.

U returns to its release height after $2u/g=2\ \mathrm{s}$ with velocity $-u$. Its remaining flight starts from exactly the height and velocity that D had at release, so its remaining duration is $T_D$. Mass does not enter the ideal free-fall equations; differing air resistance would be a change of model.

(d) Symmetry applies between equal heights.

$$t_{\mathrm{up}}=\frac ug=1\ \mathrm{s},\qquad t_{\mathrm{down}}=\frac{\sqrt{u^2+2gH}}g=\sqrt7\ \mathrm{s}>t_{\mathrm{up}}.$$

U rises for $u/g=1\ \mathrm{s}$ through $u^2/(2g)=5\ \mathrm{m}$. It falls from its top height of 35 m to the ground, taking $T_U-1=\sqrt7\ \mathrm{s}$. Equal times apply to ascent and descent between the release level and the top, not between the top and a lower landing level.