Question

A basket ball player is standing on the floor 10.0m from the basket. The height of the basket is 3.05m, and he shoots the ball at a 40 degrees angle with the horizontal from a height of 2.00m
(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?

Answer 1

(a) Acceleration of the basketball at the highest point in its trajectory is equal to the

acceleration due to gravity(g) = 9.8 ms^-2 (downward)

(b) Let the velocity of basketball = $v$ m/s

Horizontal component of velocity = $v\times cos40^{\omicron}$

Vertical component of velocity = $v\times sin40^{\omicron}$

Time taken(t) by the basketball to move 10 m horizontally = $\dfrac{10}{v\times cos40^{\omicron}}$

At this time the vertical displacement, (s) of ball will be $3.05-2=1.05 m$

$s=ut+\dfrac{1}{2}at^{2}$

$\Rightarrow1.05 =v\times sin40^{\omicron}\dfrac{10}{v\times cos40^{\omicron}}-\dfrac{1}{2}\times9.8\times\dfrac{100}{v^{2}\times cos^240^{\omicron}}$

$\Rightarrow1.05=10\times tan40^{\omicron}-\dfrac{490}{v^2cos^240^{\omicron}}$

$\Rightarrow v^2=\dfrac{490}{(10\times tan40^{\omicron}-1.05)(cos^240^{\omicron})}$

$\Rightarrow v=10.665 m/s$