Question

A basketball player is standing on the floor $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{m}}$from the basket as in Figure. The height of the basket is $\mathbf{3}\mathbf{.}\mathbf{05}\mathbf{}\mathbf{\text{m}}$, and he shoots the ball at a ${\mathbf{45}}^{\mathbf{\circ }}$angle with the horizontal from a height of$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{m}}$. (a) What is the acceleration of the basketball at the highest point in its trajectory? (b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?

Short answer

(a) The acceleration of the basketball at the highest point of its trajectory is $-9.8\mathrm{m}/{\mathrm{s}}^2$.

(b) The speed with which the player throws the ball is $10.7\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given data.

Given info: The distance between the player and the basket is $10.0\text{m}$, the height of the basket is $3.05\text{m}$, the angle made by the basketball with the horizontal is $45\circ$and the height of the player is $2.0\text{m}$.

Acceleration due to gravity.

The expression for the acceleration due to gravity in case of vertical motion is given by,

${\mathit{a}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathit{g}$

Here,

${\mathit{a}}_{\mathbf{y}}$is the vertical component of the acceleration.

(a) Determine the acceleration of the basket ball at the highest point of the trajectory.

At the highest point of trajectory, the horizontal component of the acceleration is 0 and the vertical component of the acceleration is constant and this is the acceleration due to gravity.

$a_{\mathrm{y}}=-g$

Here,

$a_{\mathrm{y}}$is the vertical component of the acceleration.

Substitute$9.8\mathrm{m}/{\mathrm{s}}^2$for$g$in the above equation.

$a_y=-9.8\mathrm{m}/{\mathrm{s}}^2$

Therefore, the acceleration of the basketball at the highest point of its trajectory is $-9.8\mathrm{m}/{\mathrm{s}}^2$.

(b) Determine the speed with which player throw the ball.

The horizontal component of the initial velocity is,

${\left(v_{\mathrm{i}}\right)}_{\mathrm{x}}=v_{\mathrm{i}}\cos \left(40.0°\right)$

Here,

$v_{\mathrm{i}}$is the initial velocity of the ball.

${\left(v_{\mathrm{i}}\right)}_{\mathrm{x}}$is the horizontal component of the initial velocity.

The vertical component of the initial velocity is,

${\left(v_i\right)}_{\mathrm{y}}=v_i\sin \left(40.0°\right)$

Here,

${\left(v_i\right)}_{\mathrm{y}}$is the vertical component of the initial velocity.

The formula to calculate the time taken by the ball to cover the horizontal distance is,

$R={\left(v_{\mathrm{i}}\right)}_{\mathrm{x}}t$

Here,

$R$is the horizontal distance covered by the ball.

$t$is the time taken by the ball.

Substitute$v_{\mathrm{i}}\cos \left(40.0°\right)$for${\left(v_i\right)}_x$and$10\text{m}$for$R$in the above equation.

$\begin{array}{c}10.0\mathrm{m}=\left(v_{\mathrm{i}}\cos \left(40.0°\right)\right)t\\ t=\frac{10.0\mathrm{m}}{v_{\mathrm{i}}\cos \left(40.0°\right)}\\ =\frac{10.0\mathrm{m}}{0.766v_{\mathrm{i}}}\\ =\frac{13.055\mathrm{m}}{v_{\mathrm{i}}}\end{array}$

Thus, the time taken by the ball to cover the horizontal distance is$\frac{13.055\mathrm{m}}{v_{\mathbf{i}}}$.

From the given figure the total horizontal distance covered by the ball is,

$\begin{array}{c}y=3.05\mathrm{m}-2\mathrm{m}\\ =1.05\mathrm{m}\end{array}$

Thus, the total horizontal distance covered by the ball is$1.05\text{m}$.

The kinematics equation for the vertical distance of the ball is,

$y={\left(v_{\mathrm{i}}\right)}_{\mathrm{y}}t+\frac{1}{2}g{t}^2$

Here,

$g$is the acceleration due to gravity.

Substitute $1.05\text{m}$for $y$,$v_{\mathrm{i}}\sin \left(40.0°\right)$for ${\left(v_{\mathrm{i}}\right)}_{\mathrm{y}},\frac{13.055\mathrm{m}}{v_{\mathrm{i}}}$for $t$and $-9.8\mathrm{m}/{\mathrm{s}}^2$for $g$in the above equation,

$\begin{array}{c}1.05\mathrm{m}=\left(v_{\mathrm{i}}\sin \left(40.0\right)\right)\left(\frac{13.055\mathrm{m}}{v_{\mathrm{i}}}\right)+\frac{1}{2}\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(\frac{13.055\mathrm{m}}{v_{\mathrm{i}}}\right)}^2\\ \left(4.9\mathrm{m}/{\mathrm{s}}^2\right){\left(\frac{13.055\mathrm{m}}{v_{\mathrm{i}}}\right)}^2=8.392\mathrm{m}-1.05\mathrm{m}\\ {\left(\frac{13.055\mathrm{m}}{v_{\mathrm{i}}}\right)}^2=\frac{7.342\mathrm{m}}{4.9\mathrm{m}/{\mathrm{s}}^2}\\ \frac{13.055\mathrm{m}}{v_{\mathrm{i}}}=\sqrt{\frac{7.342\mathrm{m}}{4.9\mathrm{m}/{\mathrm{s}}^2}}\end{array}$

Further solve the above equation.

$\begin{array}{c}\frac{13.055\mathrm{m}}{v_{\mathrm{i}}}=1.224\mathrm{s}\\ v_{\mathrm{i}}=\frac{13.055\mathrm{m}}{1.224\mathrm{s}}\\ v_{\mathrm{i}}=10.665\mathrm{m}/\mathrm{s}\\ \approx 10.7\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the time interval during which the ball is in motion is $10.7\mathrm{m}/\mathrm{s}$.