Question

A projectile is thrown from a point $O$ on the ground at an angle ${45}^{\circ }$ from the vertical and with a speed $5\sqrt{2}\text{m/s}$. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, $0.5\text{s}$ after the splitting. The other part, $t$ seconds after the splitting, falls to the ground at a distance $x$ meters from the point $O.$ The acceleration due to gravity $g=10{\text{m/s}}^2$. The value of $t$ is

Answer 1

Given, $u=5\sqrt{2}\text{m/s},$
angle of projection $\theta ={45}^{\circ }$
So, we have $u_x=u\cos \theta =5\text{m/s}$ and $u_y=u\sin \theta =5\text{m/s}$
We know that,
range, $R=\frac{2u_x{u}_y}{g}=\frac{2\times 5\times 5}{10}=5\text{m}$
Time of flight, $T=\frac{2u_y}{g}=\frac{2\times 5}{10}=1\text{sec}$
$\because$ Time of motion of one part falling vertically downwards is $=0.5\text{sec}=\frac{T}{2}$, so, time of motion of other part will be same as the component of velocity in vertical direction is same i.e. zero (Split happens at topmost point of projectile)
Hence, $t=0.5\text{sec}$