Projectile Motion Questions: Solving for Time and Velocity

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Homework Help Overview

The discussion revolves around projectile motion, specifically focusing on calculating time and velocity related to a kangaroo's jump and a ball's free fall. Participants are exploring the implications of height and distance in relation to time intervals and velocities.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to relate the time of ascent and descent for the kangaroo's jump, questioning the validity of their assumptions. They are also discussing the calculation of time intervals between images of a falling ball, with some confusion about the initial conditions and the relevance of distance measurements.

Discussion Status

There is an ongoing exploration of different interpretations of the problems presented. Some participants are questioning their initial assumptions about time and distance, while others are suggesting the use of kinematic equations to clarify their reasoning. No consensus has been reached yet.

Contextual Notes

Participants are working under the assumption that air resistance is negligible and are grappling with the implications of scale factors in their calculations. There is also uncertainty regarding the correct interpretation of the time intervals related to the stroboscope images.

liquidblot
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Q1

A kangaroo was seen to jump to a vertical height of 2.8m. for how long was it in the air?

Attempt on the solution

not possible? what do you guys think?

Q2

A ball was photographed stroboscopically as it was released and fell freely. eight images appear on the phto which is reduced in scale by a factor of five. use the fact that the acceleration of the ball is 9.8ms(-2) to discover the time interval between the flashes of the stroboscope.


Attempt on the solution

part 1. calculating the distance between the first and second image, then mutiply by 5 (due to the scale factor).

since we are taking the first and second image, it would have a overall time of 1s.

Part 2.
take the calculated length as 0.04m:

v=d/t

v= 0.04/1
= 0.04 ms(-1)

then i got stuck on this step...
 
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liquidblot said:
Q1

A kangaroo was seen to jump to a vertical height of 2.8m. for how long was it in the air?

Attempt on the solution

not possible? what do you guys think?

If you drop an object from 2.8m, how log will it take before it hits the ground?

If you can find that, then double it, because the time it took for the kangaroo to reach 2.8m is the same as the time it takes for something to fall that distance, assuming air resistance is neglected.
 
liquidblot said:
since we are taking the first and second image, it would have a overall time of 1s.


I don't know how did you find 1s.If the distance between them is 1s,then it must be the time interval between slashes of stroboscope.

liquidblot said:
Part 2.
take the calculated length as 0.04m:

v=d/t

v= 0.04/1
= 0.04 ms(-1)

then i got stuck on this step...

If d=0.04 m is true,

the equation of motion is

y=vt-\frac{1}{2}gt^2

so the distance between slashes should be

\Delta y=\Delta vt+v\Delta t-gt\Delta t

\Delta v=g\Delta t

\Delta y=gt\Delta t

No need to calculate velocity.Note that velocity is not uniform.
 
@Bill Foster
oh now i get it, so i have to use of the SUVAT equations to find the time, am i right? (^^)

@azatkgz

i think i had a mistake to take the time as 1s. and also, wouldn't i be taking the initial velocity (u) as 0 ms(-1) because the starting velocity is 0??.

and also, did i mistake the question to find the time interval between the flashes of the stroboscope to be a question on the time taken to reach from 1 image to another?

or is it surpose to mean that i have to find the displacement?
 

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