Motion Under Gravity: Physics Basics

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The discussion focuses on solving a physics problem involving two particles, P and Q, where P is dropped and Q is projected downward with an initial velocity. Participants emphasize the importance of showing calculations and understanding the initial conditions to find the time and distance each particle travels before reaching the ground. There is confusion regarding how Q catches up to P, with suggestions to derive equations for their motion and analyze their velocities at specific times. The conversation highlights the need for clarity in interpreting the problem and applying the correct equations of motion. Overall, the thread underscores the significance of detailed problem-solving steps in physics.
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Homework Statement
A particle P is let fall vertically from a height to the ground. Another particle Q is projected down vertically down 3s later with the velocity 3m/s from the same height. If both particles reach the ground simultaneously, calculate the;
1) time by each particle to reach the ground
2) distance travelled by each particle to ground
Relevant Equations
Newtons equations of motion
.
 
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What have you tried so far?
 
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GFREY said:
nothing that worked

please show what you did so we can point you in the right direction.
Also read the guidelines.
 
drmalawi said:
please show what you did so we can point you in the right direction.
Also read the guidelines.
I tried to find the initial velocity of Q using the first law of motion and from there calculate the time for Q
 
GFREY said:
I tried to find the initial velocity of Q using the first law of motion and from there calculate the time for Q
Please show us the math you tried.
 
berkeman said:
Please show us the math you tried.
v = u + gt v = 0
u = gt
u = 10m/s^2 * 3
u = 30m/s
 
GFREY said:
v = u + gt v = 0
u = gt
u = 10m/s^2 * 3
u = 30m/s

But you are given the initial velocity of ball Q in the problem.

Have you tried ##s = s_0 + v_0 t + \dfrac{at^2}{2}## formula? Can you figure out why it could be relevant here given your initial conditions?
 
drmalawi said:
But you are given the initial velocity of ball Q in the problem
drmalawi said:
But you are given the initial velocity of ball Q in the problem.

Have you tried ##s = s_0 + v_0 t + \dfrac{at^2}{2}## formula? Can you figure out why it could be relevant here given your initial conditions?
I tried it and s was 99m given conditions. Idk
 
  • #10
GFREY said:
Homework Statement:: A particle P is let fall vertically from a height to the ground. Another particle Q is projected down vertically down 3s later with the velocity 3m/s from the same height. If both particles reach the ground simultaneously, calculate the;
1) time by each particle to reach the ground
2) distance traveled by each particle to ground
Relevant Equations:: Newtons equations of motion

.
Where did you find this problem?
 
  • #11
PeroK said:
Where did you find this problem?
in my textbook. Is it wrong?
 
  • #12
Question has no viable answer. You should be able to prove that by trying to figure it out in the first place.

Does the textbook give an answer ?
 
  • #13
GFREY said:
in my textbook. Is it wrong?
It's not clear to me how Q catches P. That said , your equations should tell you that.
 
  • #14
hmmm27 said:
Question has no viable answer. You should be able to prove that by trying to figure it out in the first place.

Does the textbook give an answer ?
yes. I don't really understand the requirements of the question
 
  • #15
PeroK said:
It's not clear to me how Q catches P. That said , your equations should tell you that.
maybe it has a higher initial velocity
 
  • #16
GFREY said:
yes. I don't really understand the requirements of the question
The requirements of this forum would be that you show your work in trying to get an answer right up to the bit where it doesn't make any sense ; alternatively, explain why it has no real answer.
 
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  • #17
hmmm27 said:
The requirements of this forum would be that you show your work in trying to get an answer right up to the bit where it doesn't make any sense ; alternatively explain why it has no real answer.
okay
 
  • #18
Hint: at the time (t=3 s) ball Q is dropped, what is the velocity of ball P?
What is the velocity of ball Q and P at say t = 3,1 seconds?
 
  • #19
drmalawi said:
Hint: at the time (t=3 s) ball Q is dropped, what is the velocity of ball P?
More precisely, ball Q is projected down with initial speed 3 m/s. Sometimes "dropped" is interpreted as equivalent to "released from rest".

To @GFREY : I will ask what has already been asked but differently: when both balls are in the air, how does the vertical distance between them change? Does it increase, decrease or say the same? Please justify your choice using equations or reasoning.
 
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  • #20
PeroK said:
It's not clear to me how Q catches P. That said , your equations should tell you that.
I agree. The projected time at which the two trajectories intersect will not be positive.

It should fairly easy to write down the two requisite equations and solve them for ##t## (for the first particle to reach the ground) and ##d## (the displacement of both particles when they hit the ground together).

The first step in solving the problem should be defining the variables that will be used. The second step should be writing down equations using those variables. The third step should be solving those equations.

As @kuruman hints, a formula for the separation between the particles as a function of time (past 3 seconds anyway) should be do-able. That is not exactly the approach I'd have taken, but it is equivalent.
 
  • #21
jbriggs444 said:
As @kuruman hints, a formula for the separation between the particles as a function of time (past 3 seconds anyway) should be do-able. That is not exactly the approach I'd have taken, but it is equivalent.
Actually in my approach a formula is not necessarily necessary if one considers that, in the non-inertial frame, the relative velocity is conserved.
 
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  • #22
kuruman said:
Actually in my approach a formula is not necessarily necessary if one considers that, in the non-inertial frame, the relative velocity is conserved.
In the inertial frame as well.
 
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