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, $a=9.8m/s^{-2}$ downwards always

b) the components of the initial velocity are;

$v_{xi}=v_icos40^o, and$ $v_{iy}=v_isin40^o$

the time for the ball to move 10.0 m horizontally is;

$t=\Delta x /v_{ix} =10/(v_icos40^0)$

he vertical displacement of the ball is;

$\Delta y=3.05m -2m=1.05m$

$\Delta y=v_{iy}t+1/2a_yt^2$ becomes

$1.05m=v_isin40^0 \frac{10}{v_{i}cos40^{0}}+\frac{1}{2}(-9.8)\times \frac{10^{2}}{v_{i}(cos40^{0})^{2}}$

$v_{i}=10.7m/s^2$