Question

A particle is projected at an angle ${60}^{\circ }$ with horizontal with a speed $v=20m/s$. Taking $g=10$ $m/s^2.$ Find the time after which the speed of the particle remains half of its initial speed.

Answer 1

Given : $v=20m/s$

$\therefore$ $u_x=vcos{60}^o=20\times 0.5=10m/s$ (which remains the constant all the time)

Also $u_y=vsin{60}^o=20\times 0.866=17.32m/s$ (which decreases with time)

Let the time be t when the final velocity is half of initial velocity i.e $V_f=\frac{20}{2}=10m/s$

Thus at time $t$, the y component of the velocity is zero i.e $V_y=0$

$\therefore$ $0=u_y-gt$

OR $t=\frac{u_y}{g}=\frac{17.32}{10}=1.732$ s