Question

From the top of 25m high tower, a particle is thrown with a velocity of 5m/s at an angle 30 degree below horizontal. Find i) the time after which it hits the ground ii) horizontal distance from the base of the tower (g=10 m/s 2 ).

Answer 1

H = ut + 1/2gt^2

25 = 5t + 5t^2

5 = t + t^2

t = 1.8 s

ii) x = (5 cos 30° )1.8

= 4.5√3