A cannon ball is fired horizontally with a velocity of 50m/s from the top of a cliff 90m high,after how many seconds will it strike the plane at the foot of the cliff.at what distance from the foot of the cliff will it strike and with what velocity will it strike the ground.

Answer 1

In the vertical direction, the ball is falling under gravity and we know the following:
V0 = 0 m/sec
a = 9.81 m/s
H = 90 m

we can apply the following velocity equation:
Vf^2 = V0^2 + 2 * a * H
Vf^2 = 0 + 2 * 9.81 * 90
=1765.8
Vf = Sqrt(1765.8) = 42.02 m/s
We could apply the velocity equation:
Vf = V0 + at
42 = 0 + 9.81 * t
t = 42/9.81 =4.28 seconds
i.e. the ball hits the ground after 4.28 seconds during which it would have travelled a horizontal distance given by:

distance = V(horizontal) * t
=50 * 4.28 = 214 m
The velocity would be the resultant of both the vertical velocity (42 m/s) and horizontal velocity (50 m/s)

V = sqrt(50^2 + 42^2)= 65.3 m/sec