Question: A basketball player grabbing a rebound jump76.0 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top$\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$of this jump and (b) in the bottom 15.0 cm? (The player seems to hang in the air at the top.)

Maximum height to which player can jump is 0.76 m.

Acceleration due to gravity g=-9.8m/s

At maximum height velocity

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Also it involves the concept of free fall. The time of ascent and time of descent can be found using the concept of free fall and kinematic equations.

Formulae:

${{\mathrm{V}}_{\mathrm{f}}}^2={{\mathrm{V}}_0}^2+2\mathrm{a}∆\mathrm{y}$

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at}$

From above conditions, we can find initial velocity of player by using third kinematic equation,

${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{a\Delta y}$

$0={\mathrm{v}}_0^2+2\left(-9.8\right)\left(0.76\right)$

V₀=3.86 m/s

Total height of player’s jump is 0.76 m

We need to find total time to the top 0.15 m, for ascent and descent.

That means we need to find time at 0.61 m from base.

(0.76 m - 0.15 m = 0.61 m)

Now, consider the system from height 0.61 m to maximum height that is 0.76 m while player is going in upward direction.

We can calculate the velocity of player at 0.61 m height, by using third kinematic equation.

${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{as}$

$$\begin{gathered} {{\mathrm{V}}_{\mathrm{f}}}^2=3.{86}^2+2\left(-9.8\right)\left(0.61\right) \\ {\mathrm{V}}_{\mathrm{f}}=1.7156\mathrm{m}/\mathrm{s} \end{gathered}$$

This is the velocity at 0.61 m height with which player is ascending toward maximum height.

Now, consider the journey of player from(max. height).

Now, we can find time required to travel that journey. By using first kinematic equation.

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at}$

0=1.7156+(-9.8)t

t=0.175 s

This is the time of player going upward to top 0.15 m. But we need total time ascent and descent in top 0.15m.

So, total time spent at the top 0.15 m is 2x0.175=0.350 s

So, we know initial velocity of player is 3.86 m/s

From that we can find its velocity at 0.15 m in ascent. By using third kinematic equation.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{as} \\ {\mathrm{v}}_{\mathrm{f}}^2=3.{86}^2+2\left(-9.8\right)\left(0.15\right) \\ {\mathrm{v}}_{\mathrm{f}}^{}=3.46\mathrm{m}/\mathrm{s} \end{gathered}$$

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at}$

3.46=3.86+(-9.8)t

t=0.041 s

This is time for only ascent of 0.15 m and we need total time for ascent and descent.

So, total time 2x0.041=0.0822 s.