A cannon ball is fired horizontally with a velocity of 50 metre per sec from the top of a cliff 90m high. After how many seconds will it strike the plain at the foot of the cliff? with what velocity will it strike the ground?

First you calculate how long it will take for the object to fall the 90 meters. We'll take Earth's gravity acceleration as 9.8m/s²

Equation:

x = x₀ + v₀ * (t - t₀) + 1/2 * a * (t - t₀) ²

x - > Final position in meters

x₀ - > Starting position

v₀ - > Starting velocity

t - > Final time

t₀ - > Starting time

a - > Acceleration

Our case:

x = 0

x₀ = 90m

v₀ = 0m/s

t = ?

t₀ = 0

a = - 9.8m/s² (Minus 9.8 because it's on top of a cliff (90m) and it's falling to ground level (0m), so it's substracting)

0 = 90 - 1/2 * 9.8 * (t - 0) ²

0 = 90 - 4.9t²

4.9t² = 90

t² = 90/4.9

t = √ (18.36) = 4.28 seconds

To calculate the speed, just multiply the seconds it takes to hit the ground by the acceleration of Earth's gravity:

4.28 * 9.8 = 42m/s