# A basketball player rides a motorcycle off a ramp. When he is in midair, his teammate throws a ball straight up to him and he catches it to slam dunk. How far away from the bottom of the ramp should the teammate stand?

## To perform an amazing trick shot, a basketball player rides a motorcycle off a ramp $5.50 m$ above the ground at a velocity of 18m/s[48^@above the horizontal]. When he is in midair, his teammate throws a ball straight up to him and he catches it to slam dunk. The teammate is able to throw the ball at a speed of 19.4m/s[up] from $1.8 m$ above the ground and throws it $1.2 s$ after she leaves the ramp. How far away from the bottom of the ramp should the teammate stand?

Apr 8, 2016

$7.28 m$ rounded to two decimal places.

#### Explanation:

Approximate trajectory of the rider and point of catching the ball.

The kinematic equations start time ${t}_{\circ} = 0$ when rider leaves the ramp. Let origin of coordinate system coincide with the base of ramp where the rider leaves it. Up as well as forward are is positive directions.
Let $h$ be the height and $d$ distance from the origin when the ball is caught slam dunk.

The velocity of the rider can be resolved into its $x$ and $y$ components. As both these are orthogonal to each other therefore both can be treated independently.

Let the ball be caught after $t$ seconds after rider leaves the ramp.

Height attained by the rider in this time is found using
$h = u t + \frac{1}{2} g {t}^{2}$.
Since height of ramp when rider leaps off is $5.50 m$. Actual height traversed by rider during time $t$ is $\left(h - 5.50\right) m$. Also $g = 9.8 m {s}^{-} 2$
$\therefore h - 5.50 = {u}_{y} t + \frac{1}{2} \left(- 9.8\right) {t}^{2}$
$\implies h - 5.50 = {u}_{y} t - 4.9 {t}^{2}$ .....(1)
Negative sign of $g$ indicates that gravity is acting in a direction opposite to the direction of motion or in $- y$ direction.

Similarly for the ball thrown by the teammate, discounting for late throwing of the ball by $1.2 s$ and from height $1.8 m$ above ground.
$h - 1.8 = 19.4 t + \frac{1}{2} \left(- 9.8\right) {\left(t - 1.2\right)}^{2}$
or $h - 1.8 = 19.4 t - 4.9 \left({t}^{2} - 2.4 t + 1.44\right)$
or $h - 1.8 = 19.4 t - 4.9 {t}^{2} + 11.76 t - 7.056$ .....(2)

Subtracting (1) from (2)
$h - 1.8 - \left(h - 5.50\right) = 19.4 t - 4.9 {t}^{2} + 11.76 t - 7.056 - \left({u}_{y} t - 4.9 {t}^{2}\right)$

$- 1.8 + 5.50 = 19.4 t - {v}_{y} t + 11.76 t - 7.056$, rearranging, inserting value of ${v}_{y}$ and solving for $t$
$11.76 t + 19.4 t - 18 \times \sin {48}^{\circ} t = - 1.8 + 5.50 + 7.056$

$t = 0.605 s$ rounded to three decimal places.

To catch the ball the rider should travel a distance $d$ from origin.
$d = {v}_{x} \times t$
$d = 18 \times \cos {48}^{\circ} \times 0.605$
$d = 7.28 m$ rounded to two decimal places.