Question

A body is projected from height of $60m$ with a velocity $10m{s}^{-1}$ at angle ${30}^{\circ }$ to horizontal. The time of flight of the body is $[g=10m{s}^{-2}]$

• A. $1s$
• B. $2s$
• C. $3s$
• D. $4s$

Answer 1

The correct option is D $4s$

$h=60m$ (downward) $(+ve)$, $u_y=10\sin {30}^o$

(initial vertical velocity)

time of light will be time to reach ground i.e. to cover

$60m\frac{displacement}{\uparrow -ve}$ & $\downarrow +ve$

Put $h=60m,u\to 4u=-\frac{10}{2}=-5m/s$

$9=10m/s^2$

We get

$60=-5t+5t^2$ or $t^2-t-12=0$ or $t^2-2t+3t-12=0$ or $t(t-4)+3(t-4)=0$

$\Rightarrow t=4sec$