Question

A ball is projected horizontally. After $3s$ from projection, its velocity becomes $1.25$ times the velocity of the 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

$40m/s$

Step 1: Given Data

Time $t=3s$

Let the acceleration due to gravity be $g=9.8m/s^2$

Let the initial velocity be $u$.

Let the final velocity be $v=1.25v_x$, where $v_x$ is the velocity along $x-$axis.

Step 2: Calculate velocity along $y-$axis

For projectile motion, velocity along $y-$axis,

$v_y=u_y+a_{y}t$

For projectile motion, the initial velocity along $y-$axis is zero.

$\therefore v_y=0+9.8\times 3$

$\Rightarrow v_y=29.4m/s$

Step 3: Calculate velocity along $x-$axis

Therefore, according to Pythagoras theorem,

$v=\sqrt{{v_x}^2+{v_y}^2}$

$\Rightarrow v=\sqrt{{v_x}^2+{29.4}^2}$

$\Rightarrow {\left(1.25v_x\right)}^2={v_x}^2+{29.4}^2$

$\Rightarrow 1.5625{v_x}^2={v_x}^2+{29.4}^2$

$\Rightarrow \left(0.5625\right){v_x}^2={29.4}^2$

$\Rightarrow {v_x}^2=1539.8$

$\Rightarrow v_x\approx 40m/s$

Hence, the correct answer is option (D).