Question

A basketball player throws the ball at a 47 angle above the horizontal to a hoop which is located a horizontal distance L = 5.0 from the point of release and at a height h = 0.8 above it. What is the required speed if the basketball is to reach the hoop?

Answer 1

Answer:

$v_0 =$1.71

Explanation:

the parabolic movment is described by the following equation:

$y = tan(a)x-\frac{1}{2v_0^2(cos(a))^2}gx^2$

where y is the height of the ball, a is the angle of launch, $v_0$ the initial velocity, g the gravity and x is the horizontal distance of the ball.

So, if we want that the ball reach the hood, we will replace values on the equation as:

$0.8 = tan(47)(5)-\frac{1}{2v_0^2(cos(47))^2}(9.8)(5)^2$

Finally, solving for $v_0$, we get:

$v_0=\sqrt{\frac{-9.8(5)^2}{(0.8-tan(47)(5))2cos^2(47)}}$

$v_0 =$1.71