Question

An object is launched at a velocity of 20 m/s in a direction making an angle of 22° upward with the horizontal.What is the maximum height achieved by the projectile?

Answer 1

Given data:

Initial velocity,

`u=20\text{ m/s}`

Angle of projection,

`\theta=22\degree`

The maximum height achieved by the projectile is given as,

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

Here, g is the acceleration due to gravity.

Substituting all known values,

`\begin{gathered} H=\frac{(20\text{ m/s})^2*\sin ^2(22\degree)}{2*(9.8\text{ m/s}^2)} \\ \approx2.86\text{ m} \end{gathered}`

Therefore, the maximum height achieved by the projectile is 2.86 m.