Question

An elastic loaded balloon launcher fires balloons at an angle of (38.0° N of E) from the surfaceof the ground. If the initial velocity is 25.0 m/s, find how far away the balloons are from thelauncher when they hit the level ground again. [61.8 m]

Answer 1

Given data:

Initial velocity of the projectile;

`u=25.0\text{ m/s}`

Launch angle;

`\theta=38.0\degree`

The range of the projectile is given as,

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

Here, g is the acceleration due to gravity.

Substituting all known values,

`\begin{gathered} R=\frac{(25\text{ m/s})^2*\sin (2*38\degree)}{(9.81\text{ m/s}^2)} \\ \approx61.8\text{ m} \end{gathered}`

Therefore, the balloon will hit the ground 61.8 m away from the launcher.