Free Fall — Practice Problem Solutions

Using $g=9.8\text{ m/s}^2$.

1. $t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2(19.6)}{9.8}}=\sqrt{4}=2\text{ s}$. $v=gt=(9.8)(2)=19.6\text{ m/s}$.

2. $t_{up}=\dfrac{v_i}{g}=\dfrac{9.8}{9.8}=1\text{ s}$. Max height $=\dfrac{v_i^2}{2g}=\dfrac{9.8^2}{19.6}=4.9\text{ m}$.

3. $t=\sqrt{\dfrac{2(78.4)}{9.8}}=\sqrt{16}=4\text{ s}$. $v=(9.8)(4)=39.2\text{ m/s}$.

4. (a) $t_{up}=\dfrac{24.5}{9.8}=2.5\text{ s}$. (b) Max height $=\dfrac{24.5^2}{19.6}=30.625\text{ m}$. (c) Total flight time (up and back down to the same height) $=2t_{up}=5\text{ s}$.

5. With $\Delta x=0$: $v^2=v_i^2+2a\Delta x=v_i^2$, so $v=\pm14.7\text{ m/s}$. Physically, the ball is now moving downward, so $v=-14.7\text{ m/s}$ — same speed it was launched with, opposite direction.

6. a) Extra height gained above the rooftop $=\dfrac{v_i^2}{2g}=\dfrac{19.6^2}{19.6}=19.6\text{ m}$. Max height above the ground $=24.5+19.6=44.1\text{ m}$. b) Taking up as positive, with $v_i=+19.6\text{ m/s}$, $a=-9.8\text{ m/s}^2$, and net displacement $\Delta x=-24.5\text{ m}$ (ending 24.5 m below the launch point): $v^2=19.6^2+2(-9.8)(-24.5)=384.16+480.2=864.36\implies v=-29.4\text{ m/s}$ (i.e., $29.4\text{ m/s}$ downward at impact). c) The "time up equals time down" shortcut only works when the ball returns to the exact height it was launched from. Here it keeps falling past that point all the way to the ground, 24.5 m below the rooftop — so the "down" portion of the trip covers more distance (and takes more time) than the "up" portion, breaking the symmetry.

7. Both balls hit the ground at the exact same time. Free-fall acceleration $g$ doesn't depend on mass, and neither do the free-fall equations ($t=\sqrt{2h/g}$, etc.) — mass cancels out completely, so a heavier and lighter object dropped from the same height fall (and land) identically, ignoring air resistance.