# Free fall question

1. Jun 12, 2007

### hiddenlife5009

1. The problem statement, all variables and given/known data

In part (a) of this question suppose a stone is thrown verically upward with a speed of 12.9 m/s from the edge of a cliff and that h = 82 m. At what speed (in m/s) does the stone hit the ground?

2. Relevant equations

T = v/g
Free fall equation
V = (V^2 - 2gvt . sin(feta) + (g^2) . (t^2))^0.5

3. The attempt at a solution

By using the first equation, I have worked out how long it takes to reach max. Then by using the answer for that, and using the known values, I have used the free fall equation to find out how long it takes to reach the bottom of the cliff. Then by using the last equation I ended up with 52.956 m/s as the answer.

I was told this was not the answer and was wondering what steps I am missing or completely doing wrong.

The question may sound silly but I guess what its trying to ask is 'what is the velocity of the stone right before it hits the ground?'.

Any help would be most appreciated.

2. Jun 12, 2007

### prasannapakkiam

wow! You do not need to use such a complicated equation!

The rock is thrown up. It would soon change direction to start coming down. Calculate this height by using v^2=u^2-2as (by solving for s, also v=0 obviously).

Then add this calculated s to 85. Then use the same equation to figure out v. If you need further help ask away...:)

3. Jun 12, 2007

### hiddenlife5009

Sorry, but where are you getting 85 from?

4. Jun 12, 2007

### prasannapakkiam

Okay, there is a misunderstanding, may you please rephrase this question? I thought, h was the height of the cliff

5. Jun 12, 2007

### hiddenlife5009

Ah, height is part of the question, but I guess you meant 82, you wrote down 85. And what do you mean by add s (seconds I presume) to 82 (height)?

6. Jun 12, 2007

### prasannapakkiam

Oh sorry, Anyway, can you see how it works?

7. Jun 12, 2007

### hiddenlife5009

Not really. For starters, u = height?, and could you please refer to my previous post for the second bit I don't understand.

8. Jun 12, 2007

### prasannapakkiam

Okay, one of the 4 basic equations of motion is:
v^2=u^2+2as
v=final velocity
u=initial velocity
a=acceleration
s=distance

Now we want distance.
so we solve for s.
therefore: s=(v^2-u^2)/(2g)

Okay. When the ball reaches its maximal height, its velocity is 0 (Once you can accept that any projectile question becomes a poece of cake). So that is why we set:
v=0
g=-9.8 (the object is DECELERATING due to the effect of gravity...)
u=12.9

Okay do you get it so far? So work out s and find the distance from the ground to the maximal height of the rock (not from the cliff).

9. Jun 12, 2007

### prasannapakkiam

I see that you have used time. Try my method using distance.

10. Jun 12, 2007

### hiddenlife5009

Ok, so using that equation, I have gotten distance = 8.49m

Is that the maximum height the stone reaches? If so, where do I go from here to obtain the speed of the stone just before it hits the ground?

11. Jun 12, 2007

### prasannapakkiam

Good. So the distance from the ground (not cliff) is 8.49+h = 90.49

Okay. Now use v^2=u^2+2as
Here v is what we want to find.
Now is there an initial velocity? No. It is 0.
a=+9.8 (the stone/rock is ACCELERATING)
u=0
s=90.49
well I think the answer can be found now.

So remember acceleration whether it is a minus or plus. And try to find the easier equation to use in the context.

12. Jun 12, 2007

### hiddenlife5009

So therefore,

v = (2x9.8x90.49)^0.5

= 42.11 m/s

I'm fairly certain that is the answer. Thankyou very much for being so patient with me and helping me out, its my first time doing physics.

On a side note, there is another I need a tiny bit of help on. Got the first part of this question, but the second part I just can't grasp.

Part A

A cheetah, the fastest of all land animals over a short distance, can accelerate from zero to 18.0 m/s in three strides and to a full speed of 31.3 m/s in seconds. Assuming the first three strides are each 4.6 m long and that acceleration is constant until the cheetah reaches full speed, what is the cheetah's acceleration (in m/s2)?

Part B

If in part (a) of this question the cheetah's first three strides were each 3.7m long, how many seconds would it take to reach its full speed?

I keep saying to myself the equation I should be using is S = D/t
S = velocity/speed
D = distance/displacement
t = time

When it says the strides are now 3.7m long, I'm guessing a new acceleration is required, so I've come up with 14.77 m/s2. The answer is 2.14 seconds, but I just can't work out the solution. (Answer can be within .1).

13. Jun 12, 2007

### prasannapakkiam

That is wierd. Show how you got the answer to part a