Question

1. what is the maximum height of a softball thrown at 11.0 m/s and 50.0° above the horizontal? (think about your givens at maximum height)

Explanation:

Step1: Find vertical - initial velocity

The initial velocity $v_0 = 11.0$ m/s and the angle $\theta=50.0^{\circ}$. The vertical - component of the initial velocity is given by $v_{0y}=v_0\sin\theta$. So, $v_{0y}=11.0\times\sin(50.0^{\circ})\approx11.0\times0.766 = 8.426$ m/s.

Step2: Use kinematic equation for vertical motion

At the maximum height, the vertical velocity $v_y = 0$. The kinematic equation $v_y^{2}-v_{0y}^{2}=-2gh$ (where $g = 9.8$ m/s² and $h$ is the height). Rearranging for $h$, we get $h=\frac{v_{0y}^{2}-v_y^{2}}{2g}$. Substituting $v_y = 0$ and $v_{0y}\approx8.426$ m/s and $g = 9.8$ m/s², we have $h=\frac{(8.426)^{2}-0^{2}}{2\times9.8}=\frac{70.99}{19.6}\approx3.62$ m.

Answer:

3.62 m