Kvant Physics Problem 52
A ball is thrown vertically upward from a point that will also be taken as the point to which it later returns.
Verified: yes
Verdicts: PASS + PASS
Solve time: 11m12s
Source on kvant.digital
Problem
F52. A ball is thrown vertically upward. Which is greater: the time of ascent or the time of descent? (VIII)
Setup and Assumptions
A ball is thrown vertically upward from a point that will also be taken as the point to which it later returns. The quantity to be compared is the time of ascent, $t_{\uparrow}$, and the time of descent, $t_{\downarrow}$.
Let the initial upward velocity be $v_0$ in $\mathrm{m/s}$. The acceleration due to gravity is constant and equal to $g$ in $\mathrm{m/s^2}$, directed downward.
The motion is analyzed in an inertial reference frame fixed to the Earth. Air resistance, wind, buoyancy, and all other non-gravitational forces are neglected. The gravitational field is assumed uniform over the height reached by the ball.
Physical Principles
For motion with constant acceleration, the velocity satisfies
$$v = v_0 - gt .$$
The displacement from the launch point is
$$y = v_0 t - \frac{gt^2}{2}.$$
The highest point of the trajectory is reached when the velocity becomes zero.
Because the ball returns to its starting point, the total displacement for the entire flight is zero.
Derivation
During the ascent the velocity decreases from $v_0$ to zero. At the highest point,
$$v=0.$$
Using the velocity equation,
$$0=v_0-g t_{\uparrow},$$
which gives
$$t_{\uparrow}=\frac{v_0}{g}.$$
To find the total flight time, set the displacement equal to zero:
$$0=v_0 t-\frac{gt^2}{2}.$$
Factoring out $t$,
$$t\left(v_0-\frac{gt}{2}\right)=0.$$
One solution is $t=0$, corresponding to the launch instant. The second solution gives the time when the ball returns to the starting point:
$$v_0-\frac{gt}{2}=0,$$
hence
$$t_{\text{total}}=\frac{2v_0}{g}.$$
The descent time is the difference between the total flight time and the ascent time:
$$t_{\downarrow} = t_{\text{total}}-t_{\uparrow} = \frac{2v_0}{g}-\frac{v_0}{g} = \frac{v_0}{g}.$$
Thus
$$t_{\downarrow}=t_{\uparrow}.$$
Result
The symbolic result is
$$t_{\uparrow}=t_{\downarrow}=\frac{v_0}{g}.$$
No numerical values are given in the problem. The comparison is therefore
$$\boxed{t_{\uparrow}=t_{\downarrow}}.$$
Both times are equal, provided the ball returns to the launch point and air resistance is neglected.
Sanity Checks
The expression $v_0/g$ has dimensions
$$\frac{\mathrm{m/s}}{\mathrm{m/s^2}} = \mathrm{s},$$
so the result has the correct unit of time.
If the initial speed is doubled, both ascent and descent times double. This agrees with the fact that a larger initial speed allows the ball to rise higher while the gravitational acceleration remains unchanged.
The equality of the two times follows from the symmetry of motion under constant acceleration. At any height, the speed during descent has the same magnitude as the speed during ascent at that same height, with opposite sign.
The step most susceptible to a sign error is the use of
$$v=v_0-gt.$$
Choosing the opposite sign convention without changing all equations consistently would lead to an incorrect value for the ascent time and could falsely suggest unequal ascent and descent times.