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 the initial(g=10m/s^2)

Answer 1

The initial speed is ${\upsilon }_0=10$. The horizontal component of velocity is

${\upsilon }_{0}cos{60}^0=0.5{\upsilon }_0=5$

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

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

${0=\upsilon }_{0}cos{60}^0=-gt$

$t=\frac{10\left(\frac{\sqrt{3}}{2}\right)}{10}$

$t=0.866sec$

The required time is $t=0.866$ sec