Question

A lemming running 2.87 meters per second runs off of horizontal cliff and lands in the water 5.32 M away from the base of the cliff how high was the cliff

Answer 1

Answer: 16.8 m

Explanation:

The motion of the lemming is a projectile motion, consisting of two independent motions:

- on the horizontal (x) axis, a uniform motion, with constant speed v = 2.87 m/s

- on the vertical (y) axis, an accelerated motion, with constant acceleration $g=-9.81 m/s^2$ (downward)

The lemming lands in the water at x = 5.32 m away from the bas of the cliff, so we can calculate the time it takes to hit the water:

$t=\frac{x}{v}=\frac{5.32 m}{2.87 m/s}=1.85 s$

And now, by considering the motion on the vertical direction:

$y(t) = h +\frac{1}{2}gt^2$

we can find the height of the cliff (h) by requiring that y(t)=0:

$0=h+\frac{1}{2}gt^2\\h=-\frac{1}{2}gt^2=-\frac{1}{2}(9.81 m/s^2)(1.85 s)^2=16.8 m$