Question

A rocket is shot at32.5 m/s at a 28.8° angle,and hits a log on flat ground.How far away was the log?(Unit = m)Enter

Answer 1

Given data

The speed of the rocket is v = 32.5 m/s

The angle of projection of the rocket is theta = 28.8 degree

The expression for the distance of the log on the flat ground is given as:

`R=(v^2\sin 2\theta)/(g)`

Substitute the value in the above equation.

`\begin{gathered} R=\frac{(32.5\text{ m/s})^2*\sin (2*28.8\circ)}{9.8m/s^2} \\ R=91\text{ m} \end{gathered}`

Thus, the distance of the log on the flat ground is 91 m.