Question

A ball thrown up is caught by the thrower after 4s. how high did it go and with what velocity was it thrown ?

Answer 1

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

Answer 2

Answer:

a = -g

v= u+ at

0= u -g(2)

u= 2g = 2(10) = 20m/sec

Initial velocity of ball is 20m/sec.

and because v2-u2 = 2as

0 - (20)2 = -2(10)h

h = 20m

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