Answer to Question #109410 in Physics for Steven

Question

A player from team A is awarded 2 free throws. On the first attempt, he released the ball with the velocity of 6.0 m/s at 45° above the horizontal from a height of 2.0 m from the floor. (a) Is this a miss? (b) For the second attempt, the player shoots at 45° from the same point. What must the initial velocity of the ball so that it will go through the basket? The basket is 4.21 m from the free throw line and 3.05 m above the floor.

Answer 1

a)

$H_{max}=\frac{v^2\sin^2{45}}{2g}+h$

$H_{max}=2+\frac{6^2\sin^2{45}}{2(9.81)}=2.92\ m$

It is less than the hieght of the basket of 3.05 m. So, this is a miss.

b)

$d=R=\frac{v^2\sin{(2\cdot45)}}{g}=\frac{v^2}{g}$

$4.21=\frac{v^2}{9.81}$

$v=6.43\frac{m}{s}$