First slide

Projection Under uniform Acceleration

Question

At a certain height a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of 10 ms^-1. The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular to each other is (g=10ms^–2)

Moderate

A

1s
B

2 s
C

1.5 s
D

1.75 s

Solution

As the fragments are identical , their velocities just after explostion are equal in magnitude but oppositely directed.
Now $\large \vec V_1$ = $\large u_1 \hat i - gt \;\hat j = (u\hat i-gt\;\hat j)$
$\large and\;\vec V_2=-u_2 \hat i - gt \;\hat j = (-u\hat i-gt\;\hat j)$
Since $\large \vec V_1\;and\;V_2$ are mutually perpendicular
$\large \vec V_1\;.\;V_2=0\Rightarrow -u^2+g^2t^2=0\Rightarrow t=u/g$
$\large \therefore t = \frac{10}{10} sec=1\;sec$

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