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 qkens at maximum height) clear all 3.6 m 7.10 m 7.2 m 8.4 m

Explanation:

Step1: Find vertical - initial velocity

The initial velocity is $v_0 = 11.0$ m/s and the launch 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 final vertical velocity $v_y = 0$. The kinematic equation $v_y^2 - v_{0y}^2=-2gh$ (where $g = 9.8$ m/s²) is used to find the maximum height $h$. Rearranging the equation 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 into the equation:
$h=\frac{(8.426)^2-0^2}{2\times9.8}=\frac{70.99}{19.6}\approx3.6$ m.

Answer:

3.6 m