Question

An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reached by the object

Answer 1

H = 34.43 m

Step-by-step explanation:

Given that,

Initial speed of the object, u = 30 m/s

The angle of projection,
`\theta=60^(\circ)`

We need to find the maximum height reached by the object. Let it is H. Using the formula for maximum height reached by the projectile.

`H=(u^2\sin^2\theta)/(2g)\\\\H=((30)^2* \sin^2(60))/(2* 9.8)\\\\H=34.43\ m`

So, the maximum height reached by the object is 34.43 m.