A ball is projected with a velocity, 10 m s - 1 , at an angle of 60 ° with the vertical direction. Its speed at the highest point of its trajectory will be

Question

Aicharya · Accepted Answer

The speed of a projectile at highest point is equal to the component of velocity of projection along the horizontal direction. The angle given with the vertical is $60 °$ therefore, the angle with the horizontal will be $30 °$ .

Therefore,

$V = u \cos θ$ $\Rightarrow V = 10 \cos 30 ° = 10 \times \frac{\sqrt{3}}{2}$

$\Rightarrow V = 5 \sqrt{3} m s^{- 1}$

A ball is projected with a velocity, 10 m s - 1 , at an angle of 60 ° with the vertical direction. Its speed…

Solution