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Question

A cyclist pedals along a raised horizontal track. At the end of the track, he travels horizontally into the air and onto a track that is velocity 2.0m lower.
The cyclist travels a horizontal distance of 6.0m in the air. Air resistance is negligible. What is the horizontal velocity v of the cyclist at the end of the light track?
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A
6.3ms1
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B
9.4ms1
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C
9.9ms1
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D
15ms1
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Solution

The correct option is B 9.4ms1
Consider the problem,
This is a projectile motion question So, the horizontal and vertical component of the motion can be considered separately. The quantity that can relate both motion is the time t.

Initially, since the cyclist is on a horizontal track he has only a horizontal velocity of v, and no vertical velocity.

Air resistance is negligible, so the only force (causing an acceleration) that effects the motion is the force of gravity (which is downwards).

Consider the vertical component
Initial velocity, u=0
Acceleration,a=9.81ms2(duetogravity)Distance,s=2.0mTime,t=?(timetoreachthelowerhorizontaltrack)

s=ut+12at22=0+12×9.81×t2t=2×29.81=0.64s

So, The cyclist takes 0.64s to fall a vertical distance of 2.0m. During this 0.64s he also moves a horizontal distance of 6.0m simultaneously.

Consider the Horizontal component
Acceleration,a=0initialspeed=finalspeed=VDistancetravelled=6.0mTimetaken=0.64s

SpeedV=distancetime=6.00.64=9.4ms1

Hence, Option B is the correct answer.



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