Question
a body is thrown up with a velocity 40 m/s. At the same time another body is dropped from a height 40m. Their relative acceleration after 1.3 is
please give me the solution for the above problem
Read 4 Solution.
ADERIBIGBE DAMOLA 10 year ago
is this solution helpfull: 5 48
Here;
u=40m/s
x=40m
t=1.3s
Now,
v = u + at
as final vel. at top will be 0m/s, v=0m/s
0 = 40 - 9.8 x t
at g=9.8 & acting opposite to motion,
t=4.08s
4.08 > 1.3
hence , the motion continues till 1.3 s
now for falling body,
accelration g=9.8 m/s2
relative acceleration = acceleration for downward motion + acceleration on upward motion
a = 9.8 + 9.8
hence a=19.6 m/s2
OMKAR DEORUKHKAR 10 year ago
is this solution helpfull: 12 55
45
THEVA THAJENDRAN 10 year ago
is this solution helpfull: 5 35
right 19.7
THEVA THAJENDRAN 10 year ago
is this solution helpfull: 5 29