[Mechanics] Solving equations with 2 variables

[tigerd12]

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.

 

[HallsofIvy]

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.

 

[Architeuthis Dux]

Dear student,

Thank you for reaching out for help with this problem. Solving equations with multiple variables can be tricky, but with some practice and understanding of the concepts, you will be able to solve them successfully.

In this case, we have two different situations - one where the stone is dropped from the top of the tower, and the other where the stone is thrown downwards with a given velocity. The key here is to recognize that both stones will hit the ground at the same time, so the time taken for each stone to reach the ground will be the same.

To solve this problem, we can use the equation s = ut + 1/2at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. We can set up this equation for both stones and equate them to find the value of t.

For the first stone:

s = ? (height of the tower)
u = 0
a = 9.8 (acceleration due to gravity)
t = ?

For the second stone:

s = 0 (since it starts at the top of the tower)
u = 14
a = 9.8
t = ?

Now, equating these two equations, we get:

0 = 0 + 14t - 1/2(9.8)t^2

This simplifies to:

4.9t^2 - 14t = 0

We can factor out t from this equation:

t(4.9t - 14) = 0

This gives us two solutions: t = 0 (which we can disregard) and t = 14/4.9 = 2.86 seconds.

Since we know that the stones hit the ground at the same time, we can substitute this value of t into either equation to find the height of the tower:

s = 0 + 14(2.86) - 1/2(9.8)(2.86)^2 = 41.89 meters.

Therefore, the height of the tower is 41.89 meters.

I hope this helps you understand how to solve equations with multiple variables. Keep practicing and you will become more comfortable with these types of problems. If you have any further questions, please don't hesitate to ask.

Best of luck with your studies!

Sincerely,