Question

A ball is thrown up is caught by the thrower after 4 s. How high does it go and with what velocity

it was thrown? Take g= 10m/s2

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Answer 1

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Solution -

4 second here indicates the total time taken by the ball to reach highest point and come back to the thrower. So the highest point is reached at half the time i.e. 2 seconds. take g=9.8m/s^2

The time taken to move upward=time

he time taken to move upward=timetaken to move downwad.

So the time taken by the ball to come

down from the highest point=2 s

So from 2 nd equation of motuion,as it

was moving down:

O+1/2 gt2=h

h 5x2x2-20 m

so ut-1/2 g t2-20 while moving up

2u-20-20

u-20m/s

ie the distance it covered in t=1 sec from the highest point:h-5t2=5m

Hope it's helpful to you....

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