Question

A projectile’s launch speed is five times its speed at maximum height. Find launch angle${}^{{\mathit{\theta }}_{\mathbf{0}}}$.

Short answer

The launch angle${}^{{\theta }_0=78.5°}$

Step-by-step solution

Given information

It is given that, launch speed of the projectile is five times than its speed at maximum height.

Determining the concept of projectile motion

This problem deals with the projectile motion of an object. The projectile motion of an object is the motion of an object thrown into the air, subject to the acceleration due to gravity.Using the standard formula for the velocity of the projectile along x axis, the launch angle of the projectile can be found.

Formula:

${\theta }_x=v_{0x}\cos {}_0$

(i)

To find the launch angleθ0

Velocities at maximum height are

$$\begin{gathered} v_y=0\frac{m}{s} \\ {v}_x=v \end{gathered}$$

With equation (i),

As $v_{0x}=5v_x$

$$\begin{gathered} {\theta }_x=5v_x\cos {}_0 \\ {\theta }_0={\cos }^1\left(\frac{1}{5}\right) \\ {\theta }_0=78.5° \end{gathered}$$