a ball dropped from a height H and takes time T to reach the ground find the height of the ball after time T/3
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H=u*T+1/2*10*T^2
Here u means initial velocity which is 0 in this case
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Here u means initial velocity which is 0 in this case
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If it is helpful then plzz mark as the brainliest
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as the ball is dropped so u = 0
we want the height of ball at T/3 therefore t=T/3
a=g
therefore by using s=u*t+1/2a*t^2
for initial case when t=T,
s=1/2*g*T^2=H
for second case
s=1/2*g*T^2/9=H/9
the answer is H/9
we want the height of ball at T/3 therefore t=T/3
a=g
therefore by using s=u*t+1/2a*t^2
for initial case when t=T,
s=1/2*g*T^2=H
for second case
s=1/2*g*T^2/9=H/9
the answer is H/9
Rudrajyoti5869:
why are you using s and a you should use h and g
ok
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