Answer to Question #255750 in Mechanics | Relativity for yass

Question

A basketball player shoots a ball that is at an initial height of 2 m above the ground level at an initial velocity 

of 7.5m/s directed at an angle 𝜃0 = 55° above the horizontal. The ball enters the basket hanging at a height 

of 3.05 m above the ground level. Answer those questions: 1. Sketch the situation of this problem 2. What is the flight time of the ball from the basketball player to when it enters the basket. 3. What is the maximum height reach by the ball above the ground. 4. At what angle to the horizontal did the ball enter the basket? 

Ignore air resistance

Answer 1

$$3.05=2+7.5\sin{55}t-0.5(9.8)t^2\\t=0.2\ s\\ H=2+\frac{(7.5\sin{55})^2}{2(9.8)}=3.93\ m$$