Hard Numerical Problem???????????????
Asked by anupam | 13th Oct, 2008, 07:58: PM
Expert Answer:
For the 1st stone:
u = 0m/s
s=49m
a=g=9.8m/s2
s=ut+(1/2)at2 → 49 = 0+(1/2) x 9.8 x t2
t = 
(10) m/s = 3.16s
For 2nd stone:
t = 3.16-1 = 2.16s
a= g = 9.8m/s2
s=49m
s=ut+(1/2)at2 = 2.16u+(1/2) x 9.8 x (2.16)2
49 - 22.88 = 2.16u
26.117/2.16 = u
u = 12.09m/s
Hence,The stone was thrown at a speed of 12.09 m/s .
Answered by | 22nd Oct, 2008, 01:26: PM
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