# Freefall

• October 22nd 2007, 09:59 AM
vesperka
Freefall
Last physics problem for the day I promise :)

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3.1s later with an initial speed of 60.76m/s. They hit the ground at the same time.

The acceleration of gravity is 9.8m/s/s.

a) How long does it take the first stone to hit the ground? Answer in units of s.

b) How high is the cliff? Answer in units of m.

Based on the information given it definitely seems possible to solve this problem, but I just can't seem to put the information together.
• October 22nd 2007, 10:15 AM
mathmonster
firstly form 2 separate equations s = ut + (a(t^2))/2, then equate them

ut + (a(t^2))/2 = ut + (a(t^2))/2

taking one t as t = -3.1 and the other as just t.
• October 22nd 2007, 10:19 AM
vesperka
Hmm, I just created two position functions for each rock. The first rock is simply 4.9x^2 and the second rock is (4.9x^2)+(60.76x). I'm trying to find when their distances are equal because they hit the ground at the same time, but my calculator is only giving me 0,0 for an intersection :p

Bleh, I'm not factoring time into my equation of the second rock, which is why they never have an equal distance :\

Using your help I got t = 1.55 seconds, I think it works :D I did it backwards though, making the first rock's equation t+3.1.
• October 22nd 2007, 10:20 AM
mathmonster
which will give a value for t, then to find height of the cliff plug t into the equation in my first post.
god i havent done that for 3 years, i used to love the mechanic part of maths until i got to university
• October 22nd 2007, 10:33 AM
topsquark
Quote:

Originally Posted by vesperka
Last physics problem for the day I promise :)

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3.1s later with an initial speed of 60.76m/s. They hit the ground at the same time.

The acceleration of gravity is 9.8m/s/s.

a) How long does it take the first stone to hit the ground? Answer in units of s.

b) How high is the cliff? Answer in units of m.

Based on the information given it definitely seems possible to solve this problem, but I just can't seem to put the information together.

As always, set up your coordinate system. I have the origin at the base of the cliff and +y upward. I am defining the starting time t = 0 s when the second stone is thrown.

So for the first stone we have
$y_ 0 = h = ?$
$v_0 = 0~m/s$
$a = -9.8~m/s^2$

Giving a position equation of
$y = y_0 + v_0t + \frac{1}{2}at^2 = h - 4.9t^2$

For the second stone we have
$y_0 = h = ?$
$v_0 = -60.76~m/s$
$a = -9.8~m/s^2$

Giving position equation of
$y = y_0 + v_0(t + 3.1) + \frac{1}{2}a(t + 3.1)^2 = h - 60.76t - 4.9t^2$ (Recall that the second stone is thrown at 3.1 s.

They hit the ground at the same time, so at time t they both have a position y = 0. So
$h - 4.9(t + 3.1)^2 = h - 60.76t - 4.9t^2$

Giving t = 1.55 s.

So how tall is the cliff? Pick a stone:
$0 = h - 4.9(1.55 + 3.1)^2$

Giving h = 105.95 m.

-Dan
• October 22nd 2007, 04:51 PM
mathmonster
see i got the height of the cliff as 11.7m when i first did the calculation, but it seemed wrong so i plugged t into the equation for the stone with initial velocity 60.76m/s and got the height of the cliff as 106m, which makes more sense as the stone takes 1.55s to hit the floor and the initial vel. is 61m/s.
• October 22nd 2007, 08:07 PM
topsquark
Quote:

Originally Posted by mathmonster
see i got the height of the cliff as 11.7m when i first did the calculation, but it seemed wrong so i plugged t into the equation for the stone with initial velocity 60.76m/s and got the height of the cliff as 106m, which makes more sense as the stone takes 1.55s to hit the floor and the initial vel. is 61m/s.

I'm not doing so hot today, am I? (Puke)

I forgot to add the 3.1 s when using that equation. If you plug the 1.55 s into the "thrown rock" equation you get the correct height. If you plug 1.55 + 3.1 (s) into the "dropped rock" equation you also get the right answer.

I have corrected this in my original post. (And I also got the 105.95 m answer.)