# Vertical Projectile Motion, Calculate height

1. Oct 24, 2013

### TheRedDevil18

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

A body is dropped from a height of 300 m. At exactly the same moment
another body is projected from the ground vertically upwards with a
velocity of 150 m.s^-1.

Calculate the …
3.2.1 time it will take for the two objects to reach the same height.

3. The attempt at a solution

delta x = vi*t + 1/2*g*t^2

for falling body:
delta x = 1/2*g*t^2.............1

for projected body:
300 - delta x = 150t - 1/2*g*t^2..........2

1+2

300 = 150t
t = 2's

Okay, so I know I will have to solve this simultaneously. I cheated a bit and got this answer from the memo, but I dont understand the second equation, which is the part about the 300 - deltax. Isn't delta x the same for both?, or is it because one is thrown up and the other down which is why they are not the same?, can somebody please explain the second equation to me, especially the 300 - deltax?

2. Oct 24, 2013

### tiny-tim

Hi TheRedDevil18!

x1 = 1/2*g*t^2.............1

x2 = 150t - 1/2*g*t^2..........2

when they meet, x1 = ∆x, and x2 = … ?

3. Oct 24, 2013

### TheRedDevil18

Ok, I see, In order for x1 to be equal to x2, x2 would have to be 300 - Δx. So this is the height and not the displacement, right?. I mean it would be wrong to say that x1 = x2 because the displacement of x1 is not the same as the displacement of x2?. What about the signs?, do I not worry about the signs?

4. Oct 24, 2013

### tiny-tim

Hi TheRedDevil18!
all correct

your standard equation assumes the position starts at 0, ie it is the displacement

we have to adjust the equation so that ∆x means the same thing
You must always always ALWAYS worry about the signs!

In this case, one of your displacements is measured upwards, and one downwards

if you had measured both upwards, the 300 - ∆x would instead be … ?

5. Oct 24, 2013

### TheRedDevil18

So if I take down as positive then shouldnt the second equation be 300 - Δx = -150t + 1/2*g*t^2, because initial velocity is upwards, so negative, and gravity is downwards, so positive?

6. Oct 24, 2013

### tiny-tim

(but wouldn't that give a different result? )

if both x1 and x2 are measured in the same direction ,

then (in terms of x1 and x2) you must have 300 = … ?

7. Oct 24, 2013

### TheRedDevil18

If it where measured in the same direction then both displacements would be positive so it would be 300 = x1 + x2, but in this situation they are in opposite directions.

This is what I am confused about, If I take down as positive, then:

x1 = 1/2gt^2.............1

For projectile going up (negative)

-x2(going up) = -150t(thrown up) + 1/2gt^2...............2, gravity should be positive (down)

then 300 = 1/2gt^2 - 150t + 1/2gt^2

I know it doesnt make sense when I calculate the answer, but to me the directions do make sense or am I wrong?

8. Oct 24, 2013

### haruspex

No, that would be measuring them in opposite directions, each being 0 at time 0.
No, you don't want the minus sign in front of the x2. If x2 is measured as +ve down then it will take negative values. The RHS above will give you that. Putting the minus on the left as well will give +ve values for x2.

9. Oct 25, 2013

### TheRedDevil18

Looking at those two equations it seems that for the falling body they took down as positive and for the projected body they took up as positve, because on the right hand side, their is a negative for gravity

10. Oct 25, 2013

### haruspex

Yes, and the delta x in each case is the distance the upper body falls.
Btw, the problem can be solved very simply by noting that the two have the same acceleration. What does that imply about the relative velocity?

11. Oct 25, 2013

### TheRedDevil18

Not sure, but maybe when they reach the same height their velocity relative to each other is 0 m/s?. For the second equation why can't I choose down as positive?. If I do it would effect the outcome.

12. Oct 25, 2013

### tiny-tim

Hi TheRedDevil18!
No, of course not … one is still going up, and the other is coming down fast!

Think … when they reach the same height, what is their relative velocity? (use haruspex's hint)

13. Oct 25, 2013

### TheRedDevil18

I think it would be 150m/s, which is equal to the initial velocity of the body thrown up?

14. Oct 25, 2013

### tiny-tim

that's correct

the relative acceleration is 0, so the relative speed is constant, ie 150 m/s !

does that give you a short-cut to finding the time?​

15. Oct 25, 2013

### TheRedDevil18

Im not sure how, maybe if I say that (Velocity of falling body) + (Velocity of body thrown up) = 150

And for this equation:
x2 = 150t - 1/2*g*t^2

What if I chose down as positive?, they seem to have chosen up as positive but down wouldnt work out, why?. Is it because x2 would be negative?

Last edited: Oct 25, 2013
16. Oct 25, 2013

### haruspex

No, we're dealing with relative velocity, i.e. the difference between them (assuming they're measured in the same direction). They start 300m apart and the relative velocity is 150m/s. So how long will it take to close the gap?

17. Oct 26, 2013

### TheRedDevil18

t = d/s
= 300/150
= 2's

But what about acceleration?, how can it be zero when it is still under the influence of gravity?

18. Oct 26, 2013

### tiny-tim

yes
this is relative acceleration

x1'' = -a
x2'' = -a

so x1'' - x2'' = 0

sp x1' - x2' = … ?

19. Oct 26, 2013

### TheRedDevil18

Oh, I see where this is going,

x1' - x2'
= gt - (150 -gt)
= -150 m/s

Which is the relative velocity. Would this work for any problem, lets say, a ball was thrown up from a 300m high cliff and a second ball launched from the ground 1 second later at 50m/s. Calculate the time at which they meet?

20. Oct 26, 2013

### tiny-tim

yes, the relative velocity stays the same