First slide

Projection Under uniform Acceleration

Question

A body is projected from corner A of a hall of length 16 m and height 4 m with velocity u at an angle $α$ with horizontal such that it just touches the roof and hits the other-end B of the floor. Then ‘u’ is equal to

Moderate

A

$4 \sqrt{5} m / s$
B

$10 m / s$
C

$4 \sqrt{10} m/s$
D

$8 m / s$

Solution

$tan α = \frac{4 H}{R} = \frac{4 \times 4}{16} \Rightarrow α = 45^{o}$
Now, $\frac{u^{2} \sin^{2} 45^{o}}{2 \times 10} = 4 \Rightarrow u = 4 \sqrt{10} m / s$

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