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

Step-by-step 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=(x)/(v)=(5.32 m)/(2.87 m/s)=1.85 s`

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

`y(t) = h +(1)/(2)gt^2`

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

`0=h+(1)/(2)gt^2\\h=-(1)/(2)gt^2=-(1)/(2)(9.81 m/s^2)(1.85 s)^2=16.8 m`