Question

An object is launched at a speed of 20 m/s. If it rises to a height of 12 m, at what angle was it launched? ________° above the +x direction

Answer 1

Given:

The initial speed of the object is,

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

The maximum height is,

`H=12\text{ m}`

To find:

The angle of launch above the X-axis

Step-by-step explanation:

The maximum height above the X-axis is,

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

Here, the angle above the X-axis is,

`\theta`

Substituting the values we get,

`\begin{gathered} 12=((20)^2sin^2\theta)/(2*9.8) \\ sin^2\theta=(12*2*9.8)/(400) \\ sin\theta=0.7668\text{ \lparen taking positive value only\rparen} \\ \theta=50.1\degree \end{gathered}`

Hence, the angle of launch is,

`50.1\degree`