Chapter 4, Problem 24P

A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. P4.12a). 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. P4.12b) with center of mass elevations y_i = 1.20 m, y_max = 2.50 m, and y_f = 0.700 m.

Figure P4.12

To determine

The time of flight of the basketball star.

Answer to Problem 24P

The time of flight of the basketball star is $0.852\text{ s}$.

Explanation of Solution

Section 1:

To determine: The initial velocity of basketball star to go up.

Answer: The initial velocity of basketball star to go up is $4.03\text{m}/\text{s}$.

Given information:

The horizontal distance covered by the basket ball star is $2.80\text{ m}$, the centre of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

From the instant the star leaves the floor until just before he lands, the basketball star is a projectile.

The equation to calculate the upward motion of his flight is,

$v_{\text{yf}}^2=v_{\text{yi}}^2+2a\left(y_{\text{f}}-y_{\text{i}}\right)$ (I)

Substitute $0$ for $v_{\text{yf}}$, $-9.8\text{m}/{\text{s}}^2$ for $a$, $1.85\text{ m}$ for $y_{\text{f}}$ and $1.02\text{ m}$ for $y_{\text{i}}$ in above equation to find $v_{\text{yi}}$.

$\begin{array}{c}0=v_{\text{yi}}^2+2\left(-9.8\text{m}/{\text{s}}^2\right)\left(1.85\text{ m}-1.02\text{ m}\right)\\ v_{\text{yi}}=4.03\text{m}/\text{s}\end{array}$

Section 2:

To determine: The final velocity of basketball star to go up.

Answer: The initial velocity of basketball star to go up is $-4.32\text{m}/\text{s}$.

Given information:

The horizontal distance covered by the basket ball star is $2.80\text{ m}$, the centre of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

Substitute $0$ for $v_{\text{yi}}$, $-9.8\text{m}/{\text{s}}^2$ for $a$, $0.900\text{ m}$ for $y_{\text{f}}$ and $1.85\text{ m}$ for $y_{\text{i}}$ in above equation (I) to find $v_{\text{yf}}$.

$\begin{array}{c}{v}_{\text{yf}}^2=0+2\left(-9.8\text{m}/{\text{s}}^2\right)\left(0.900\text{ m}-1.85\text{ m}\right)\\ v_{\text{yf}}=-4.32\text{m}/\text{s}\end{array}$

Section 3:

To determine: The time of flight of the basketball star.

Answer: The time of flight of the basketball star is $0.852\text{ s}$.

Given information:

The horizontal distance covered by the basketball star is $2.80\text{ m}$, the center of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

The equation to calculate the hang time of basketball star is,

$v_{\text{yf}}=v_{\text{yi}}+at$

Substitute $4.03\text{m}/\text{s}$ for $v_{\text{yi}}$, $-9.8\text{m}/{\text{s}}^2$ for $a$ and $-4.32\text{m}/\text{s}$ for $v_{\text{yf}}$ in above equation to find $t$.

$\begin{array}{c}\left(-4.32\text{m}/\text{s}\right)=\left(4.03\text{m}/\text{s}\right)+\left(-9.8\text{m}/{\text{s}}^2\right)t\\ t=\frac{\left(-8.35\text{m}/\text{s}\right)}{\left(-9.8\text{m}/{\text{s}}^2\right)}\\ =0.852\text{ s}\end{array}$

Conclusion:

Therefore, the time of flight of the basketball star is $0.852\text{ s}$.

To determine

The horizontal velocity component of the basketball star at take off.

Answer to Problem 24P

The horizontal velocity component of the basketball star at take off is $3.29\text{m}/\text{s}$.

Explanation of Solution

Given information:

The horizontal distance covered by the basket ball star is $2.80\text{ m}$, the centre of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

The formula to calculate horizontal velocity component of the basketball star is,

$x=v_{\text{xi}}t$

Substitute $2.80\text{ m}$ for $x$ and $0.852\text{ s}$ for $t$ in above equation to find $v_{\text{xi}}$.

$\begin{array}{c}2.80\text{ m}=v_{\text{xi}}\left(0.852\text{ s}\right)\\ v_{\text{xi}}=\frac{2.80\text{ m}}{0.852\text{ s}}\\ =3.29\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the horizontal velocity component of the basketball star at take off is $3.29\text{m}/\text{s}$.

To determine

The vertical velocity component of the basketball star at takeoff.

Answer to Problem 24P

The vertical velocity component of the basketball star at takeoff is $4.03\text{m}/\text{s}$.

Explanation of Solution

Given information:

The horizontal distance covered by the basketball star is $2.80\text{ m}$, the centre of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

From the section 1 of part (a), the vertical component of the velocity of the basketball star at takeoff is,

$v_{\text{yi}}=4.03\text{m}/\text{s}$

Conclusion:

Therefore, the vertical velocity of the basketball star at takeoff is $4.03\text{m}/\text{s}$.

To determine

The takeoff angle.

Answer to Problem 24P

The takeoff angle is $50.8°$.

Explanation of Solution

Given information:

The horizontal distance covered by the basket ball star is $2.80\text{ m}$, the centre of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

The formula to calculate take off angle is,

$\theta ={\tan }^{-1}\left(\frac{v_{\text{yi}}}{v_{\text{xi}}}\right)$

Substitute $4.03\text{m}/\text{s}$ for $v_{\text{yi}}$ and $3.29\text{m}/\text{s}$ for $v_{\text{xi}}$ in above equation to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{4.03\text{m}/\text{s}}{3.29\text{m}/\text{s}}\right)\\ =50.8°\end{array}$

Conclusion:

Therefore, the takeoff angle is $50.8°$.

To determine

The time of flight of the deer.

Answer to Problem 24P

The time of flight of the deer is $1.12\text{ s}$.

Explanation of Solution

Section 1:

To determine: The upward velocity of deer going up.

Answer: The upward velocity of deer going up is $5.04\text{m}/\text{s}$.

Given information:

The horizontal distance covered by the basketball star is $2.80\text{ m}$, the center of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

Substitute $0$ for $v_{\text{yf}}$, $-9.8\text{m}/{\text{s}}^2$ for $a$, $2.50\text{ m}$ for $y_{\text{f}}$ and $1.20\text{ m}$ for $y_{\text{i}}$ in above equation (I) to find $v_{\text{yi}}$.

$\begin{array}{c}0=v_{\text{yi}}^2+2\left(-9.8\text{m}/{\text{s}}^2\right)\left(2.50\text{ m}-1.20\text{ m}\right)\\ v_{\text{yi}}=5.04\text{m}/\text{s}\end{array}$

Section 2:

To determine: The upward velocity of deer going down.

Answer: The downward velocity of deer going down is $-5.94\text{m}/\text{s}$.

Given information:

The horizontal distance covered by the basketball star is $2.80\text{ m}$, the center of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

Substitute $0$ for $v_{\text{yi}}$, $-9.8\text{m}/{\text{s}}^2$ for $a$, $0.700\text{ m}$ for $y_{\text{f}}$ and $2.50\text{ m}$ for $y_{\text{i}}$ in above equation (I) to find $v_{\text{yf}}$.

$\begin{array}{c}{v}_{\text{yf}}^2=0+2\left(-9.8\text{m}/{\text{s}}^2\right)\left(0.700\text{ m}-2.50\text{ m}\right)\\ v_{\text{yf}}=-5.94\text{m}/\text{s}\end{array}$

Section 3:

To determine: The time of flight of the deer.

Answer: The time of flight of the deer is $1.12\text{ s}$.

Given information:

The horizontal distance covered by the basketball star is $2.80\text{ m}$, the center of mass at the elevation of $1.02\text{ m}$, the maximum height is $1.85\text{ m}$ above the floor and at the elevation of $0.900\text{ m}$ when he touches the ground again.

The equation to calculate the hang time of deer is,

$v_{\text{yf}}=v_{\text{yi}}+at$

Substitute $5.04\text{m}/\text{s}$ for $v_{\text{yi}}$, $-9.8\text{m}/{\text{s}}^2$ for $a$ and $-5.94\text{m}/\text{s}$ for $v_{\text{yf}}$ in above equation to find $t$.

$\begin{array}{c}\left(-5.94\text{m}/\text{s}\right)=\left(5.04\text{m}/\text{s}\right)+\left(-9.8\text{m}/{\text{s}}^2\right)t\\ t=\frac{\left(-10.98\text{m}/\text{s}\right)}{\left(-9.8\text{m}/{\text{s}}^2\right)}\\ =1.12\text{ s}\end{array}$

Conclusion:

Therefore, the time of flight of deer is $1.12\text{ s}$