Motion problem with two rocks being released at different times

  • Thread starter Thread starter mjolnir80
  • Start date Start date
  • Tags Tags
    Motion Rocks
Click For Summary
SUMMARY

The discussion revolves around a physics problem involving two rocks released from a balloon ascending at 3.0 m/s. The first rock is released at 50 meters above the ground, while the second rock is thrown 2 seconds later from 56 meters. Both rocks hit the ground simultaneously, and the challenge is to determine the initial velocity of the second rock with respect to the ground. The solution employs the equation ∆d = Vi t + ½ a t², confirming that the initial velocity of the first rock is 0 m/s, while the second rock's initial velocity must account for the balloon's ascent.

PREREQUISITES
  • Understanding of kinematic equations, specifically ∆d = Vi t + ½ a t²
  • Knowledge of relative motion in physics
  • Familiarity with the concept of free fall and gravitational acceleration
  • Basic understanding of initial velocity and its implications in motion problems
NEXT STEPS
  • Study the effects of initial velocity in projectile motion
  • Learn about relative velocity concepts in physics
  • Explore advanced kinematic equations for varying acceleration
  • Investigate real-world applications of motion equations in engineering
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in solving motion problems involving multiple objects and relative velocities.

mjolnir80
Messages
54
Reaction score
0

Homework Statement


the question is like this :you are in a balloon that is rising at a rate of 3.0m/s . you have two rocks. you release one when you are 50m above the ground. 2 seconds later you throw the other rock towards the ground . the two rocks hit the ground at the same time . what was the initial velocity of the second rock (with respect to the ground)


Homework Equations


we have to solve it only using the simple motion equations such as:
∆d = Vi t + ½ a t2



The Attempt at a Solution


what i did was to first say that :

t(rock1)=t(rock2)-2s and we also know that the second rock is released when we are 56m above the ground

then i proceeded to calculate the time it takes for the first rock to hit the ground (considering the fact that Vi for the first rock is 0)

then i subtracted 2 from this time and used it to calculate Vi for the second rock
i used ∆d = Vi t + ½ a t2 to calculate both these things

the problem is I am not sure I am doing this right and i would appreciate it if someone could confirm my method
thanks in advance
 
Physics news on Phys.org
that Vi for the first rock is 0
Surely -3 m/s ? The rocks will be moving with the balloon until they are dropped or thrown.

Apart from that your method looks OK.
 
Last edited:
Mentz114 said:
Surely -3 m/s ? The rocks will be moving with the balloon until they are dropped or thrown.

Apart from that your method looks OK.

oops i was thinking about a different problem
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
40
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
5K