Question

A body is projected with a velocity of 20 m/s in a direction making an angle of ${60}^{\circ }$ with the horizontal. Calculate
(i) its position after 0.55
(ii) velocity after 0.55

Answer 1

Time of flight of the particle $T=\frac{2\times 20\times \sqrt{3}}{2\times 10}$

$T=2\sqrt{3}$ seconds

i)Position in X-direction after 0.55 second $X=10\times 0.55=5.5m$

In Y-direction ,$Y=10\sqrt{3}\times 0.55-5\times (0.55{)}^2$

$Y=8m$

position=$(5.5,8)$

ii)Velocity in X direction $=10m/s$

Velocity in Y direction $v=u-gt$

$v=10\sqrt{3}-10\times 0.55=11.8m/s$

velocity $=10\hat{i}+11.8\hat{j}$