Question

A ball is projected horizontally. After $3$ second from projection its velocity becomes $1.25$ times of the velocity of projection. Its velocity of projection is:-

• A. $10m/s$
• B. $20m/s$
• C. $30m/s$
• D. $40m/s$

Answer 1

The correct option is D $40m/s$

As no horizontal force is acting

$V_2=V+\left(0\right)t$ [acceleration is zero]

$V_2=V$

In vertical direction,

$V_1=\left(0\right)+gt$ [initial vertical component of vec 0 is]

$V_1=gt\Rightarrow V_1=10t$

At, $t=3s$

$V_1=\sqrt{V_1^2+V_1^2}=1.25V$

$\Rightarrow \sqrt{V^2+(3\times 10{)}^2}=1.25V$

$\Rightarrow (30{)}^2+V^2=(1.25{)}^2{V}^2$

$\Rightarrow (30{)}^2+V^2=(1.25{)}^2{V}^2$

$(30{)}^2=[1.5625-1]V^2$

$\Rightarrow \sqrt{\frac{900}{0.5625}}=V$

$\Rightarrow 40m/s=v$

option$D$ is correct