A Ball is thrown vertically upward
with velocity "v"& such that it
rises for T seconds (T>1)What
is distance traversed by ball
during last one second ?
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Answer. Initial velocity u= v
Final velocity v=0
u=v,v=0
g=9.8m/s^2(downwards)
g=-9.8m/s^2(upward)
2gs=v^2-u^2
2*(-9.8)*s=(0)^2-(0)^2
-19.6s=0
s=19.6/0=19.6
s=19.6m
When ball return to the surface of the earth s=0 v=u+at
0=0+9.8t0=9.8t
t=9.8/0=9.8
t=9.8s
Total time taken is 9.8+9.8=19.6s
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