a missile launches at a velocity of 30.0 m/s at an angle of 60.0° to the horizontal. what is the maximum…

Explanation:

Step1: Find vertical - initial velocity

The initial velocity is $v_0 = 30.0\ m/s$ and the launch angle $\theta=60.0^{\circ}$. The vertical - component of the initial velocity is $v_{0y}=v_0\sin\theta$. So, $v_{0y}=30\sin60^{\circ}=30\times\frac{\sqrt{3}}{2}=15\sqrt{3}\ m/s$.

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 $g = 9.8\ m/s^2$) 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 $v_{0y}=15\sqrt{3}\ m/s$ and $g = 9.8\ m/s^2$ into the equation: [ \begin{align*} h&=\frac{(15\sqrt{3})^2-0^2}{2\times9.8}\ &=\frac{15^2\times3}{2\times9.8}\ &=\frac{675}{19.6}\ &\approx34.4\ m \end{align*} ]

Answer:

34.4 m