If a ball is thrown up with a certain velocity and it attains a height of 40 m
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The question is incomplete.
However, assuming that either initial velocity u or time t is to be calculated, here is the solution.
given, h= 40m, g=-10m/s².
Applying 3rd equation of motion,
2as = v² - u² [as ball is thrown up, v = 0]
on solving, we get u= 20√2 m/s.
Putting u = 20√2 in V=u+at,
0=20√2-10t
t=10√2s
:)
However, assuming that either initial velocity u or time t is to be calculated, here is the solution.
given, h= 40m, g=-10m/s².
Applying 3rd equation of motion,
2as = v² - u² [as ball is thrown up, v = 0]
on solving, we get u= 20√2 m/s.
Putting u = 20√2 in V=u+at,
0=20√2-10t
t=10√2s
:)
Answered by
0
Answer:
40
Explanation: intial speed = 10*4
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