Basketball hoop

Classical Mechanics Level 3

A basketball hoop is at a height of $h_1 = 3.05\text{ m}$ above the floor. The center of the basket is at a distance of $L = 5.425\text{ m}$ horizontally from the free-throw line. A basketball player shoots free throws and the ball leaves his hand at the moment when its center is exactly above the free-throw line at a height of $h_2 = 2.45\text{ m}$ above the floor.

Find the minimum speed at which the player should shoot the ball so that the ball directly passes through the hoop.

Details and Assumptions:

$g = 9.81\text{ m/s}^2.$
Air resistance is negligible.

The answer is 7.7.

1 solution

Ajit Athle
Nov 24, 2017

Relevant wiki: Projectile Motion

We can use the general equation of parabolic motion: $y = x \tan θ - \dfrac{g}{2} \left( \dfrac{x}{u\cos θ}\right) ^2$ where $y$ , $x$ & $g$ are known.

$y = (3.05-2.45) \text{ m}$
$x = 5.425 \text{ m}$
$g = 9.81 \text{ m/s}^2$

We get $\dfrac{g}{2} \left( \dfrac{x}{u\cos θ}\right) ^2\ = (x \tan \theta - y)$

$u =\frac{x}{\cos \theta} \sqrt{\frac{g}{2}\frac{1}{(x \tan \theta - y)}}$

From this function, under the given conditions, $u$ is maximized at about $θ=48.2 ^\circ$ and thus the minimum initial velocity can be determined by the equation: $3.05-2.45=5.425 \tan(48.2^\circ)- \frac{9.81}{2} \times \left( \frac{5.425}{u\cos(48.2^\circ)} \right)^2$ from where $u=7.70907$ m/s

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Syed Hamza Khalid - 3 years, 6 months ago

Basketball hoop

The answer is 7.7.

1 solution

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