Vertical Projectile Motion, Calculate height

[TheRedDevil18]

Homework Statement

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.

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 don't 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?

 

[tiny-tim]

Hi TheRedDevil18!

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

your equations of motion are

x₁ = 1/2*g*t^2....1

x₂ = 150t - 1/2*g*t^2...2

when they meet, x₁ = ∆x, and x₂ = … ? ​

 

[TheRedDevil18]

Hi TheRedDevil18!


your equations of motion are

x₁ = 1/2*g*t^2....1

x₂ = 150t - 1/2*g*t^2...2

when they meet, x₁ = ∆x, and x₂ = … ? ​

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?

 

[tiny-tim]

Hi 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?.

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

What about the signs?, do I not worry about the signs?

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 … ?

 

[TheRedDevil18]

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 … ?

So if I take down as positive then shouldn't 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?

 

[tiny-tim]

So if I take down as positive then shouldn't the second equation be 300 - Δx = -150t + 1/2*g*t^2 …

(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 = … ? ​

 

[TheRedDevil18]

(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 = … ?

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 doesn't make sense when I calculate the answer, but to me the directions do make sense or am I wrong?

 

[haruspex]

If it where measured in the same direction then both displacements would be positive so it would be 300 = x1 + x2

No, that would be measuring them in opposite directions, each being 0 at time 0.

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)

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.

 

[TheRedDevil18]

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

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

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

 

[haruspex]

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

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?

 

[TheRedDevil18]

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?

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.

 

[tiny-tim]

Hi TheRedDevil18!

Not sure, but maybe when they reach the same height their velocity relative to each other is 0 m/s?.

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)

 

[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)

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

 

[tiny-tim]

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

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?​

 

[TheRedDevil18]

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

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 wouldn't work out, why?. Is it because x2 would be negative?

 

[haruspex]

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

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?

 

[TheRedDevil18]

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

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?

 

[tiny-tim]

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

yes

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

this is relative acceleration

x₁'' = -a
x₂'' = -a

so x₁'' - x₂'' = 0

sp x₁' - x₂' = … ?

 

[TheRedDevil18]

this is relative acceleration

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

so x1'' - x2'' = 0

sp x1' - x2' = … ?

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, let's 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?

 

[tiny-tim]

yes, the relative velocity stays the same

 

[haruspex]

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, let's 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?

Yes, but in this case you need to calculate the velocity of the first ball at the moment the second ball is thrown.

 

[TheRedDevil18]

Ok, thanks guys for your help