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## Homework Statement

A basketball player is awarded free-throws. The center of the basket is [itex]4.21m[/itex] from the foul line and [itex]3.05m[/itex] from the ground. On the first free throw he throws the ball at an angle of 35 degrees above the horizontal with a speed of [itex]v_0 = 4.88 m/s[/itex]. The ball is released [itex]1.83m[/itex] from the floor. What is the maximum height reached by the ball?

## Homework Equations

[itex]x = x_0 + v_0 t + (1/2) a t^2[/itex]

[itex]v^2 = v_0^2 + 2 a \Delta x[/itex]

[itex]v = v_0 + a t[/itex]

[itex]v_y^2 = v_y_o^2 + 2gh[/itex]

## The Attempt at a Solution

Since we were given both [itex]v_x[/itex] and [itex]v_y[/itex], we wanted to find the distance straight from the release point of the ball to the center of the rim.

[itex]3.05m - 1.83m = 1.22m[/itex], which is height from the player's release point to the center of the rim.

Then, [itex]cosθ = {1.22}/{x}[/itex]

Therefore, the distance is [itex]5.20m[/itex]

(This information was only relevant for a later part of the question)

Then,

[itex]v_y^2 = v_yo^2 + 2gh[/itex] where 'h' is height, which is what we needed to solve for.

Solving for [itex]h[/itex], we find that

[itex]h={-v_yo}/{2g}[/itex]. Then,

[itex]h={-(4.88 sin35)^2}/{2(-9.8m/s^2)} ≈ 0.4m[/itex]

Adding [itex]0.4m[/itex] to the basketball's original height of [itex]1.83m[/itex], and you get the ball's maximum height to be [itex]2.23m[/itex] above the ground.

There is no soution for me to verify this answer with, so I would appreciate anyone telling me where I messed up, or in the less likely scenario, that I got the answer correct.

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