Question

172. A ball is projected from the ground at an angle q with the
horizontal. After 1 second it is moving at angle 45° with
the horizontal and after 2 seconds it is moving horizontally!
What is the velocity of projection of the ball ?​

Answer 1

Answer:

10√5 m/s

Explanation:

vy = usinq - gt

vx = ucosq

tan∝ = (usinq - gt)/ ucosq

tan45° = (usinq - gt)/ ucosq

(usinq - g)/ ucosq = 1

(usinq - g) = ucosq

tan0° = (usinq - 2g )/ ucosq = 0 , usinq = 2g , ucosq = g , tanq = 2

u² = 5g²

u = g√5 = 10√5 m/s