A rock released from rest to fall on a small airless planet, find average speed?

Question

a rock released from rest to fall on a small airless planet falls y = gt^2 meters in t seconds, g is a constant. Suppose that the rock falls to the bottom of a crevasse 20 meters below and reaches the bottom in 4 seconds.

a. find the value of g

b. find the average speed for the fall

c. with what speed did the rock hit the bottom?

I'm a bit stuck. I think I know how to do the average speed, but how do i figure out what speed the rock hit the bottom?

Answer 1

a) g=20/4^2 = 1.25 m/s^2

b) average speed =20/4=5 m/s

c) speed =sqr(2*1.25*20) = 7.07 m/s



hint: need to use conservation of energy.

m*g*h=1/2*m*v^2

solve for v.

check:

h=v^2/2*g

=(7.07)^2/(2*1.25) = 19.99396 meters

or

h=Vave*4=5 m/s *4 s=20 meters

Answer 2

a. y = gt² ──► g = y/t² = 20/(4)² = 1.25 m/s²

b. v,av = Δy / Δt = 20 / 4 = 5 m/s

c. v = y' = 2gt = 2(1.25)(4) = 10 m/s

EDIT: AIP, the equation in your check (y=v²/(2g)) disagrees with the problem statement:

y = gt² ──► v = y' = 2gt ──► t = v/(2g)

y = gt² = g[v/(2g)]² = v²/(4g)

Answer 3

Here's what we know.

y=gt^2

y=20 m

t=4 s (when hits the bottom)

So you want to know g and velocity when it hits the ground

First, solve for g

g=y/(t^2)

Now, you want to know the speed. So take a derivative of y with respect to t which is,

v=2gt

and look, you know g and you know t when it hits the ground. You can take it a step further and substitute g into the equation for v

v=2(y/t^2)t= (2y)/t

Crazy! You never needed know g in the first place!