Two objects - when do they meet?

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The discussion revolves around a physics problem involving two objects moving towards each other, one thrown downward and the other propelled upward. The initial calculations for their positions were based on constant velocities, which led to an incorrect meeting height of 85.3448 m. Participants highlighted the need to incorporate gravitational acceleration into the equations to accurately model the motion of both objects. A suggested equation for free-fall was provided to help refine the approach. The conversation emphasizes the importance of correctly applying physics principles to solve the problem effectively.
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Homework Statement


An object is thrown downward with an initial
speed of 4 m/s from a height of 99 m above the
ground. At the same instant, a second object
is propelled vertically from ground level with
a speed of 25 m/s.

The acceleration of gravity is 9.8 m/s^2.

At what height above the ground will the
two objects pass each other? Answer in units
of m.

Homework Equations


None - I believe you make your own equation.
Look below to see what I mean.

The Attempt at a Solution


Well I tried to find the position equation of both
objects, then set them equal to each other to
find the time at which their positions are equal.

For the top object, since at time 0 his position
is 99 m, and his initial velocity is 4 m/s, I found
its position equation to be x(t) = -4t + 99. The
-4t is negative because its traveling down.

For the bottom object, since at the time 0 his
position is also 0, it has no constant. His initial
velocity is 25 m/s, so I got his position equation
to be x(t) = 25t. The 25 t is positive because he
is propelled upwards.

So I set the two equations equal, and solved for t:

-4t + 99 = 25t
99 = 29t
t = 3.4138

I plugged this time into one of my position
equations to find the position when they
meet:
x(t) = 25t
x(3.4138) = 25(3.4138)
x = 85.3448 m

This answer seemed to make sense, because
since the bottom object is propelled significantly
faster than the top object, they should meet
somewhere close to the 99 m mark.

But low and behold, 85.3448 m is incorrect.

Help!
 
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You quoted a figure for the acceleration of gravity but your equations are all assuming constant velocity. Put g into the problem.
 
Don't forget, your "top" and "bottom" objects won't stay at 4ms^{-1} & 25ms^{-1} forever, they experience gravity. I suggest you look at this equation for a free-fall object:

x(t) = x_0 + ut - \frac{1}{2}gt^2 See if you can apply this to the problem
 
BerryBoy said:
Don't forget, your "top" and "bottom" objects won't stay at 4ms^{-1} & 25ms^{-1} forever, they experience gravity. I suggest you look at this equation for a free-fall object:

x(t) = x_0 + ut - \frac{1}{2}gt^2 See if you can apply this to the problem

Do you think I'm giving too much help :confused:
 
BerryBoy said:
Do you think I'm giving too much help :confused:

Solving the problem for them is too much. Beyond that, how explicit with hints you want to be is completely up to you.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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