Question

A body of mass $m$ is projected from ground with speed $u$ at an angle $\theta$ with horizontal. The power delivered by gravity when it is at half of maximum height from ground is

• A. $\frac{mgu\cos \theta }{\sqrt{2}}$
• B. $\frac{mgu\sin \theta }{\sqrt{2}}$
• C. $-\frac{mgu\sin \theta }{\sqrt{2}}$
• D. Both (b) and (c)

Answer 1

The correct option is D Both (b) and (c)


$H_{max}=\frac{u^2{\sin }^2\theta }{2g}$
Applying 3rd eqn of motion at half of maximum height :
$v_y^2=(u\sin \theta {)}^2-2g\left(\frac{H_{max}}{2}\right)$
$v_y^2=(u\sin \theta {)}^2-\frac{2g{u}^2{\sin }^2\theta }{4g}$
$v_y=\pm \frac{u\sin \theta }{\sqrt{2}}$
for the two points of half the maximum height

At point A, $\overrightarrow{P}=\overrightarrow{F}.\overrightarrow{V}$
$=\left(mg\right)\left(\frac{u\sin \theta }{\sqrt{2}}\right)\cos \pi =\frac{-mgu\sin \theta }{\sqrt{2}}$
{angle between vectors is ${180}^{\circ }$}

At point B,
$\overrightarrow{P}=\overrightarrow{F}.\overrightarrow{V}$ {angle between vectors here is $0^{\circ }$}
$\overrightarrow{P}=\frac{+mgu\sin \theta }{\sqrt{2}}$