Question

A particle is projected with speed 10m/s at an angle 60 with horizontal .Then the time after which it's speed becomes half of initial (g=10m/s^2)

Answer 1

The initial speed is $v_0=10$. The horizontal component of velocity is

$v_0\cos 60^o=0.5v_0=5$.

At the highest point of the trajectory of the projectile, its speed is 5 m/s which is half the initial speed.

The time taken to reach the maximum height (when the vertical component of the speed becomes 0) is

$0=v_0\sin60^o-gt\\ t=\frac{v_0\sin60^o}{g} \\ t=\frac{10 (\sqrt{3}/2)}{10} \\ t=0.866 \;sec$

The required time is $0.866 \;sec$