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What is the speed of the rock right before it hits the ground?

  • Thread starter turtledove
  • Start date
1. Homework Statement

A boy kicks a rock off a cliff with a speed of 19.4 m/s at an angle of 52.5° above the horizontal. The rock hits the ground 5.41 s after it was kicked.

a) How high is the cliff? (60.3 m)

b) What is the speed of the rock right before it hits the ground?

c) What is the maximum height of the rock measured from the top of the cliff?

2. Homework Equations

d = Vit + 1/2 at^2
vf^2 = vi^2 + 2ad
d= vt

3. The Attempt at a Solution

a) I found this by using the equation:
d= (-19.4sin52.5)(5.41) + (0.5)(9.8)(5.41)^2
= 60.3 m, this is correct

b) I am having trouble figuring out the speed as it hits the ground. I was trying:
Vf^2 = 0 + 2(9.8)(60.3)
= 34.4 m/s, however this seems to be wrong..... what am i doing wrong?

c) I do not know how to start this part.

Somebody please help me!
 

gneill

Mentor
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There are separate y and x components of the velocity to consider. The speed at impact will be the composition of the two.

Have you considered energy conservation as a route to the answer?
 
Hmm....does that mean it is simply 19.4 m/s?
 

gneill

Mentor
20,489
2,615
Hmm....does that mean it is simply 19.4 m/s?
Nope. Not in this case, where the landing point is lower than the starting point -- it converts potential energy to kinetic energy on the way down...
 
Ok, I tried something different:

Vfx = 19.4cos52.5
= 11.8 m/s

Vfy = 19.4sin52.5 + sqrt 2(60.3)(9.8)
= 15.4 + 34.37
= 49.76 m/s

V^2 = (Vx)^2 + (Vy)^2
= 51.1 m/s

This is wrong again :(
 

verty

Homework Helper
2,157
198
Turtledove, your Vfy is totally incorrect.
 
would i just keep it as Vfy = 19sin52.5 ? what other values need to be incorporated? Im so confused with this question!
 

verty

Homework Helper
2,157
198
No, I won't answer that question. This is homework and I'm sure you know enough to work it out. You know where the problem is, so draw a sketch and do some thinking about what is incorporated.
 

gneill

Mentor
20,489
2,615
Your expression for Vfy is not correct. If you want to use energy change to figure the final y-velocity, then you should start with the initial kinetic energy (in the y-direction) and add the energy due to potential energy change before converting back to velocity.

You could also just add the potential energy change to the overall initial kinetic energy, no need to worry about components.

Another way to the final y-velocity is to consider that you're given the time of flight, so just apply v(t) = vo + a*t.
 

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