Question

A cannon ball is launched off a 100 m cliff horizontally with an initial horizontal velocity (Vx) of 50 m/s.

Answer 1

We have the equation of motion $s = ut+\frac{1}{2} at^2$ , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here displacement = 100 m, Initial velocity = 0 m/s, acceleration = 9.81 $m/s^2$

Substituting

$100 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 100 = 4.9t^2\\ \\ t = 4.52 seconds$

A cannon ball is launched off a 100 m cliff horizontally with an initial horizontal velocity (Vx) of 50 m/s will take 4.52 seconds to reach the ground.