Question

Two bodies were thrown simultaneously from the same point, one straight up, and the other at angle of $\theta ={30}^0$ to the horizontal. The initial velocity of each body is $20m{s}^{-1}$ Neglecting air resistance, the distance between the bodies at $t=1.2$ later is

• A. $20m,$
• B. $30m$
• C. $24m$
• D. $50m$

Answer 1

Answer: (C)

$24m$

Let body (1) goes straight up means making an angle of ${90}^{\circ }$
and the other body makes an angle of ${30}^{\circ }$ with the horizontal as given in the question.

Given velocity of two bodies: $u=20m/s$

The relative velocity $\left(V_R\right)$ is given as:

$V_R=\sqrt{u^2+u^2-2u\times ucos(90-\theta )}$

$V_R=\sqrt{(20{)}^2+(20{)}^2-2\left(20{)}^{2}cos\right(90-30)}$

$V_R=\sqrt{2\times (20{)}^2-2\times (20{)}^{2}Cos{60}^{\circ }}......\because cos{60}^{\circ }=\frac{1}{2};$

$V_R=\sqrt{2\times (20{)}^2-2\times (20{)}^2\times \frac{1}{2}}$;

$V_R=20\sqrt{2-1}$

$V_R=20m/s=u-m/s$

As relative velocity remains unchanged,

Using the formula $Distance\left(d\right)=Speed\left(s\right)\times Time\left(t\right)$ , the required distance covered is:

$d=20\times 1.2m$

$d=24m$

Option (C) is correct.