During a basketball game, a player shoots a ball from half-court. When the ball reaches its maximum height of $4.5 \mathrm{~m}$ above the floor, it is moving at $5.0 \mathrm{~m} / \mathrm{s}$. If the ball was released from $2.5 \mathrm{~m}$ above the floor, what was the angle above the horizontal of the ball's initial velocity?

Question

Quizplus · Accepted Answer

To find the angle, use the vertical motion equations. At maximum height, vertical velocity is 0. Using $v_y^2 = v_{0y}^2 - 2g(y - y_0)$, solve for $v_{0y}$. Then, use $v_{0y} = v_0 \sin(\theta)$ and $v_{0x} = v_0 \cos(\theta)$ with $v_{0x} = 5 \, \text{m/s}$ (since horizontal velocity remains constant) to find $\theta$. Solving gives $\theta \approx 51^\circ$.

During a Basketball Game, a Player Shoots a Ball from Half-Court