A firecracker is fired and rises to a height of 1000 m. find the (1) velocity by which it was released (2) time taken by it to reach the highest point [take g = 9.8m/s 2 ]
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Answered by
226
(1) h=1000m
v= 0 m/s (at highest point)
g= 9.8 m/s( taken to be negative according to sign convention)
Using 3rd equat. of motion,
v^2= u^2 + 2as
0 = u^2- 2*9.8*1000
-u^2= -19600
u^2= 19600
u=140 m/s
(2) u=140 m/s (as calculated above)
1st equation of motion,
v= u+at
0=140+(-9.8)t
-140= -9.8t
140= 9.8t
t= 1400/98
t= 14.3 seconds
t= 14 seconds( approximately)
v= 0 m/s (at highest point)
g= 9.8 m/s( taken to be negative according to sign convention)
Using 3rd equat. of motion,
v^2= u^2 + 2as
0 = u^2- 2*9.8*1000
-u^2= -19600
u^2= 19600
u=140 m/s
(2) u=140 m/s (as calculated above)
1st equation of motion,
v= u+at
0=140+(-9.8)t
-140= -9.8t
140= 9.8t
t= 1400/98
t= 14.3 seconds
t= 14 seconds( approximately)
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80
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