Kinematics Question: Child on Trampoline Jump Height Calculation

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Homework Help Overview

The discussion revolves around kinematics problems involving a child jumping on a trampoline and a truck decelerating. The original poster presents a question about calculating the height reached by the child based on the time spent in the air, while another problem involves determining the initial velocity of a truck before braking.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the interpretation of the total time in the air and how it relates to the time taken to reach the highest point. There is also an exploration of kinematic equations and their applicability given the information provided.

Discussion Status

Some participants have offered guidance on how to approach the problems, while others express uncertainty about the realism of the numbers used in the second problem. The conversation reflects a mix of attempts to clarify concepts and check assumptions without reaching a definitive conclusion.

Contextual Notes

There is mention of missing information, such as the mass of the child in the first problem, and concerns about the realism of the data in the second problem. Participants are encouraged to verify the problem statements for accuracy.

Gurvir
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Homework Statement


A child on a trampoline remains in the air for 1.5s after having jumped straight up. What height did the child reach?

Homework Equations


All kinematics equations
I was thinking of using Net Force but we don't have the mass and if you use it how could you find displacement.

The Attempt at a Solution



d = vit + 1/2at2
vi= 0
So:
d= 1/2at2
d = 1/2(9.81m/s2)(1.5s)2
d= 11m

The answer is suppose to be 2.8m
 
Last edited:
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The time given is the total time in the air. So what's the time it took to reach the highest point?
 
Doc Al said:
The time given is the total time in the air. So what's the time it took to reach the highest point?

Half of that which is 0.75s, and wow. There it is, simplest question. Thanks man!

Omg I keep getting stuck on these questions, kinematics is so easy but I can't remember anything. Check back I will have a new question posted in 1min.

Homework Statement


A truck uniformly decelerates by 5.00m/s2 to a velocity of 60.0m/s [W] in 30.0s. What was the truck's initial velocity before the driver stepped on the brakes?

Homework Equations


All kinematics equations
I'm looking for an equation but I don't have distance so I can't figure one out that will work. The closest formula I could maybe see working is
vf2 = vi2 + 2ad

The Attempt at a Solution


I'm not sure.

(EDIT)

Found it:

Solution

a = Δv / Δt
a = vf - vi / Δt

Rearrange it and you get this
vi = 60m/s -at

Put in the variables
vi = 60m/s - (-5.00m/s)(30.0s)
vi = 60m/s + (5.00m/s)(30.0s)

Answer:
vi = 210 m/s [W]
 
Last edited:
Looks good!
 
Based on the numbers you quote the result is correct, but then it sure is one fast truck. If I were you I would double check that the given numbers are correctly taken from the problem text.
 
Filip Larsen said:
Based on the numbers you quote the result is correct, but then it sure is one fast truck. If I were you I would double check that the given numbers are correctly taken from the problem text.
Yes, the numbers are comically unrealistic. Worth checking that you copied the problem correctly, but it's also not unusual for problems to use unrealistic data. (Unfortunately.)
 
Doc Al said:
Yes, the numbers are comically unrealistic. Worth checking that you copied the problem correctly, but it's also not unusual for problems to use unrealistic data. (Unfortunately.)

i guess I am just lucky, my textbook uses realistic numbers, it even has a textblock on the side that says they're all realistic =]..

and yes, that's an insanely fast truck xD
 

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