Question

A body is projected from the ground with the velocity of ${30ms}^{-1}$ , making an angle of ${30}^{}$ with the horizontal. Find the maximum height obtained by it. $(Takeg={10ms}^{-2})$

• A. $32.5$ m
• B. $62.5$ m
• C. $11.25$ m
• D. $12.55$ m

Answer 1

The correct option is C $11.25$ m

Given : $u=30m/s$ $\theta ={30}^o$
Maximum height $H_{max}=\frac{u^2{\sin }^2\theta }{2g}$ $(\sin 30=0.5)$
$⟹H_{max}=\frac{{30}^2\times {\sin }^2\left(30\right)}{2\times 10}=11.25m$