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

`v_0 =`1.71

Step-by-step explanation:

the parabolic movment is described by the following equation:

`y = tan(a)x-(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)-(1)/(2v_0^2(cos(47))^2)(9.8)(5)^2`

Finally, solving for
`v_0`, we get:

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

`v_0 =`1.71