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## Homework Statement

A bastketball player is 2m tall and stands 10m from the basket.The height of the basket is 3m and the player throws the ball with speed 11m/s. At what angle with respect to the horizontal should he throw the ball so that is goes through to basket?

## Homework Equations

g=gravity, h= maximum height vi= initial velocity x=angle R= maximum range

h=vi^2 * (sin(x))^2 / 2g

R= vi^2 * sin 2x /g

## The Attempt at a Solution

Here is what've done:

h=vi^2 * (sin(x))^2 / 2g

x= Inverse SIN (SQRT ( h*2g / vi^2 ))

height = 1m

x= Inverse Sin(SQRT (1*2(9.8)/ 11^2)) = 23.73 degree

R= vi^2 * sin 2x /g

x= (Inverse SIN(R*g / vi^2)) / 2

R= 10m

x= (Inverse SIN(10*9.8 / 11^2)) / 2 = 27.04degree

both angles aint the same,

is it because the height of the basket is different to the player's height?

do i need to put "h=vi^2 * (sin(x))^2 / 2g " and

"R= vi^2 * sin 2x /g " into one eqn in order to get the angle?

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