Projectile Motion Questions: Solving for Time and Velocity

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The discussion revolves around solving two projectile motion problems: the time a kangaroo is in the air after jumping to a height of 2.8m and the time interval between flashes of a stroboscopic photograph of a falling ball. For the kangaroo, it is suggested that the time to reach the peak height is equal to the time to fall back down, which can be calculated using kinematic equations. In the case of the ball, participants are attempting to determine the time interval between images based on the distance between them and the known acceleration due to gravity. There is some confusion regarding the calculations and assumptions about initial velocity and displacement, indicating a need for clarity on using kinematic equations effectively. Overall, the discussion highlights the importance of understanding projectile motion principles and the application of relevant equations.
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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?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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