Question

A projectile is fired horizontally off the top of a cliff with an initial velocity of 43 m/s. It hits the ground 3.0 seconds later.
b) How high is the cliff?

Answer 1

Answer:

44.1 m

Explanation:

The vertical motion of the projectile is a free fall motion, so it is a uniform accelerated motion. This means that we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where , taking downward as positive direction:

s is the vertical displacement

u = 0 is the initial vertical velocity of the projectile (it is fired horizontally)

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

We know that the projectile hits the ground at

t = 3.0 s

Therefore, by substituting this value into the equation, we find the vertical displacement of the projectile at that time, that is equal to the height of the cliff:

$s=0+\frac{1}{2}(9.8)(3)^2=44.1 m$