[Mechanics] Solving equations with 2 variables

AI Thread Summary
To solve the problem of two stones dropped from a tower, the first stone falls under gravity with an equation for distance as h = -4.9t². The second stone, thrown one second later with an initial velocity of 14 m/s, falls according to h = -4.9(t-1)² - 14(t-1). Both equations represent the same height, allowing for simultaneous equations to be set up. By equating the two expressions for height, the values of height and time can be determined. This approach effectively uses kinematic equations to find the height of the tower.
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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