From a cliff of 49m high, a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. Find out the speed with which he threw the second stone.PLEASE!! (Use equations of motion etc. and show the working!)............

2
by Bon

Answers

The Brainliest Answer!
2014-08-31T00:43:50+05:30
For first stone
h = 49m
g = 9.8m/s²
t = ?
u = 0
apply
h = ut + gt²/2
49 = 0 + 9.8t²/2
t² = 10
t = √10 = 3.16s
for second stone
t = 3.16 - 1 = 2.16s
h = 49m
g = 9.8m/s²
u = ?
again apply
h = ut + gt²/2
49  = u*2.16 + 9.8*2.16²/2
49 = 2.16*u + 22.86
26.14/2.16 = u
u = 12.10m/s

Thnx but again ambiguity striked my mind.........why is u not equal to 0 for the second stone?
2014-08-31T22:53:54+05:30

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