Calculating Maximum Height of a Rock Thrown Vertically Upward

[Hughey85]

A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 30 m high. When it hits the ground at the base of the cliff the rock has a speed of 25 m/s.

(a) Assuming that air resistance can be ignored, find the initial speed of the rock.

(b) Find the greatest height of the rock as measured from the base of the cliff.


I got part (a) to be 6.0 m/s, but do not know how to get part (b). If I know both velocities, what equation should I use to find max height? Thank you!

 

[maverick280857]

Please show your work.

Hint #1 for part (a): Consider the motion of the ball upwards from the projection point and back to it. Does it tell you something about the velocity of the ball as it comes back to the projection point? Is the motion symmetric?

Hint for part (b): Do you know kinematic equations involving distances and velocities? How many different kinds are available? What information do you have or don't have?

Is the velocity of the ball when it reaches the ground greater than, less than or equal to the projection velocity?

 

[Hughey85]

For part (a) I used Ui + Ki = Uf + Kf ...which gave me

mgy1 + .5mvi^2 = mgyf +.5mvf^2

I Then solved for vi


For part (b) I'm not sure what equation to use. I was thinking to use this one:

v^2 = vo^2 + 2a(delta)x

the final velocity should equal 0 right? because the rock is on the ground. What is the acceleration, a , though? Without, a , I cannot find x.

 

[Hughey85]

Nevermind...(a) would equal -9.81 because its the acceleration due to gravity. I solved for x, but then I noticed I had to add on the 30 m. of the cliff. I came up with 32 m. Thanks a lot for the help!

 

[tony873004]

Being that they gave you .26 kg as the mass of the rock, and this value is not needed in the computation, it implies that this book or teacher likes to throw trick questions at you. So... There might be another trick in this question. If the rock is thrown vertically from the top of the cliff, it's not going to drop to the base of the cliff. It's going to drop to the top of the cliff, exactly where it was thrown. It may bounce or roll from that point and go over the cliff, in which case velocity initial would = 0 for the second part of the problem. But if you want it to hit the base of the cliff from the top of the cliff, you have to throw it at an angle that is not equal to vertical.