First slide

Projection Under uniform Acceleration

Question

A ball is projected obliquely with a velocity 49 ms^–1 strikes the ground at a distance of 245 m from the point of projection. It remained in air for

Moderate

Solution

$\large R = \frac{{{u^2}\sin 2\theta }}{g} \Rightarrow 245 = \frac{{49 \times 49 \times \sin 2\theta }}{{9.8}}$
$\large \sin 2\theta = 1 = \sin {90^0}$
$\fn_phv \large 2\theta = {90^0} \Rightarrow \theta = {45^0}$
$\large \therefore T = \frac{{2u\sin \theta }}{g} = 5\sqrt 2 \sec$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App