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

Answer:

H = 34.43 m

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.

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m$

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