Distance and Displacement

Question

A person standing at the edge of the cliff kicks a stone over the edge at a speed of 20 m/s. The cliff is 70 m above the surface below. With what speed does it strike the ground below?

Answer 1

Hello Hailey!

A good thing to keep in mind is that vertical displacement, velocity, and acceleration is unrelated to horizontal displacement, velocity, and acceleration.

Basically, what this means is that, after the rock is kicked, no more horizontal acceleration acts on it, it launches at a constant horizontal speed. You don’t need more horizontal vectors to calculate the overall speed

so you basically use the displacement equation which is:

y = yi+ V•(t) + 0.5•(a)•(t)^2

where y is the final position in meters

yi is the initial position it was kicked at

V is the vertical velocity

a is acceleration which is -g in this case.

and t is time

we get:

0 = 70 + 0(t) ** there is no initial vertical velocity when it is kicked horizontally** +0.5 (-9.81) (t)^2

we get time being: 3.77s when we solve for t.

after that, we use the velocity equation using the time we got to determine the vertical speed it reached a moment before hitting the ground.:

Vf= Vi + a(t)

Vf= 0 -9.81(3.77)

Vf= - 36.98m/s or a speed of 36.98 m/s downwards (without the magnitude).

if you remember we had that constant Vx, and now we have instantaneous Vy, so we can use the pythagorean theorem to find the overall vector of speed:

Vx^2 + Vy^2 = Vz^2

Vz = sqrt (37^2 + 20^2) = 42m/s with an angle of arctan (-37/20) = -31 degrees