Question

A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. P4.24a). His motion through space can be modeled precisely as that of a particle at his center of mass, which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor and is at elevation 0.900 m when he touches down again. Determine

(a) his time of flight (his “hang time”)

(b) his horizontal and

(c) vertical velocity components at the instant of takeoff, and

(d) his takeoff angle.

(e) For comparison, determine the hang time of a whitetail deer making a jump (Fig. P4.24b) with center-of-mass elevations yi 5 1.20 m, ymax 5 2.50 m, and yf 5 0.700 m.

Short answer

1. Hang time$=0.85s$
2. Horizontal velocity component at the instant of take-off$=3.294m/s$
3. Vertical velocity component at the instant of take-off$=4.033m/s$
4. Take-off angle$=50.76°$
5. Hang time of whitetail deer$=1.12s$

Step-by-step solution

Solving for vertical component of initial velocity

Let the initial height of the basketball star be

Maximum height,

Final height,

Horizontal displacement

(c)

Taking only the vertical components

${v_y}^2-{u_y}^2=2a_y(h-h_i)$

$$\begin{gathered} 0-{u_y}^2=2(-9.8)(1.85-1.02) \\ {u_y}^2=16.268 \\ {u}_y=4.033m/s \end{gathered}$$

Hence the vertical component of initial velocity is$4.033m/s$

In second half of the path

Solving for final velocity

$$\begin{gathered} v_y{\text{'}}^2-u_y{\text{'}}^2=2a_y(h_f-h) \\ {v}_y{\text{'}}^2-0=2(9.8)(0.9-1.85) \\ {v}_y{\text{'}}^2=-18.62 \\ {v}_y\text{'}=-4.315m/s \end{gathered}$$

Solving for the hang time

(a)

Putting the values in

$$\begin{gathered} v_y\text{'}=u_y+a_{y}t \\ -4.315=4.033-9.8t \\ 9.8t=8.348 \\ t=0.85s \end{gathered}$$

Therefore, the time of flight came out to be$0.85s$

Solving for the horizontal velocity component

(b)

$$\begin{gathered} x=u_{x}t \\ 2.8=u_x(0.85) \\ {u}_x=3.294m/s \end{gathered}$$

Solving for the take-off angle

(d)

Angle between the two components of initial velocities is given by

$\tan \theta =\frac{u_y}{u_x}$

$\theta ={\tan }^{-1}\left(\frac{4.033}{3.294}\right)$

$\theta =50.76°$

Hence, the take-off angle is$50.76°$

Now for the whitetail deer

(e)

Similar approach as from step 1 to step 3, just the values will be changed. Refer to step 1 to step 3 for the formulae used.

First half of the trajectory:

$$\begin{gathered} 0-{u_y}^2=2(-9.8)(2.5-1.2) \\ -{u_y}^2=-25.48 \\ {u}_y=5.05m/s \end{gathered}$$

Second half of the trajectory:

$$\begin{gathered} {v_y}^2-0=2(9.8)(0.7-2.5) \\ {v_y}^2=-35.28 \\ {v}_y=-5.94m/s \end{gathered}$$

Hence, for the flight time,

$$\begin{gathered} v_y=u_y-gt \\ -5.94=5.05-9.8t \\ 9.8t=10.99 \\ t=1.12s \end{gathered}$$

Therefore the flight time of a whitetail deer is 1.12s