[Mechanics] Solving equations with 2 variables

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SUMMARY

The discussion focuses on solving a physics problem involving two stones dropped from a tower, where one stone is dropped and another is thrown downwards with an initial velocity of 14 m/s. The key equations derived are based on the dynamics of free fall under gravity, specifically using the formula s = -4.9t² + vt for distance fallen. The first stone falls a distance h = -4.9t², while the second stone falls a distance h = -4.9(t-1)² - 14(t-1). By equating these two equations, the height of the tower can be determined.

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tigerd12
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Hello there, I am having a lot of trouble solving a question with which I cannot even begin, I have tried making 2 equations and equation, but I just end up with some number = 0, which is of course wrong, here is the question.

QUESTION

A stone is dropped from the top of a tower. One second later another stone is thrown vertically downwards from the same point with a velocity of 14ms^-1. If they hit the ground together find the height of the tower.

I've managed to get that..

First rock

s = ?
u = 0
v = ?
a = 9.8
t = ?

Second rock

s = ?
u = 14
v = ?
a = 9.8
t = ?

Now where do I go from here? I've tried making 2 equations using the same base equation i.e. v^2 = u^2 + 2as with the different values and then equating them, which doesn't work.

Any help would be much appreciated, thanks.
 
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You haven't really shown much. Where did you get the formula v^2+ u^2+ 2as? "Conservation of energy".

I would just use the dynamics equations: If an object falls under gravity with initial velocity v, then s= -4.9t2+ vt is the distance it has fallen. The first rock has fallen, at time t, a distance h= -4.9t2 and the second rock has has fallen a distance h= -4.9(t-1)2- 14(t-1). Those are negative because it is downward. The second equation has t-1 because the rock was thrown one second after the first rock so the time it has been falling is 1 less. Since the height, h, and the time, t, are the same for those two rocks, you can solve the two equations for h and t. h is what you are asked.
 

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