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 the vertical - component of the initial velocity

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

Step2: Use the kinematic equation for vertical motion

The kinematic equation $v_y^2 = v_{0y}^2-2gh$ is used to find the maximum height $h$. At the maximum height, the vertical velocity $v_y = 0$.
We can re - arrange the equation $v_y^2 = v_{0y}^2-2gh$ for $h$: $h=\frac{v_{0y}^2 - v_y^2}{2g}$.
Since $v_y = 0$, $h=\frac{v_{0y}^2}{2g}$.
We know that $g = 9.8\ m/s^2$ and $v_{0y}\approx8.426\ m/s$.
$h=\frac{(8.426)^2}{2\times9.8}=\frac{70.99}{19.6}\approx3.62\ m$

Answer:

$3.62\ m$