A ball thrown up is caught by the thrower after 4s. How high did it go and with what velocity was it thrown? how far below its heighest point was it 3 s after start?

Asked by  | 21st Aug, 2012, 12:15: AM

Expert Answer:

the time taken to move upward=time taken to move downwad
so the time taken by the stone to come down from the highest point=2 s
so from 2 nd equation of motuion,as it was moving down: 0+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

Answered by  | 23rd Aug, 2012, 03:45: PM

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