Question

A particle $A$ is projected from the ground with an initial velocity of $10m/s$ at an angle of ${60}^{\circ }$ with horizontal. From what height h should an another particle $B$ be projected horizontally with velocity $5m/s$ so that both the particles collides at point $C$ if both are projected simultaneously $(g=10m/s^2)$

• A. $10m$
• B. $30m$
• C. $15m$
• D. $25m$

Answer 1

The correct option is C $15m$

Given,

Velocity of $1^{st}$ projectile, $u=10m{s}^{-1}$

Angle of projection, $\theta ={60}^o$

Time of flight, $t=\frac{2u\sin {60}^o}{g}=\frac{2\times 10\sin {60}^o}{10}=\sqrt{3}\sec$

Given,

Velocity of $2^{nd}$ projectile in vertical direction, $u_y=0$

Time, $t=\sqrt{3}\sec$

Height of fall, $s=u_{y}t+\frac{1}{2}g{t}^2=0+\frac{1}{2}\times 10\times {\left(\sqrt{3}\right)}^2=15m$

Hence, height is $15m$