A stone dropped from a certain height travels 2/5th of its length in the last second of its motion. find the time taken to reach the ground and the height from where it is dropped (g=10m/s^2)
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given,
u=0;
h=2/5 *h
g=10 m/s^2
so ,u=0 we get ,
h =1/2gt^2 ---------------------1
so 2/5th of the height is
2/5*h=1/2gt^2-1/2g(t-1)[t-1 because of the last second]
2/5*h=1/2g(2t-1)
2/5*h=5(2t-1)[as we now from equation 1 we get ]
2/5*1/2gt^2=5(2t-1)
1/5gt^2=5(2t-1)
2t^2=10t-5
⇒2t^2-10t+5=0(therefore by spitting middle term )
2t^2-5t-5t+5
⇒2t(t-1)+5(t-1)
⇒2t+5 and t-1 are the factors
therefore time can 1 or -5/2
as time cannot be negative we get time is 1
by this
h=1/2*10*1
h=5 m
u=0;
h=2/5 *h
g=10 m/s^2
so ,u=0 we get ,
h =1/2gt^2 ---------------------1
so 2/5th of the height is
2/5*h=1/2gt^2-1/2g(t-1)[t-1 because of the last second]
2/5*h=1/2g(2t-1)
2/5*h=5(2t-1)[as we now from equation 1 we get ]
2/5*1/2gt^2=5(2t-1)
1/5gt^2=5(2t-1)
2t^2=10t-5
⇒2t^2-10t+5=0(therefore by spitting middle term )
2t^2-5t-5t+5
⇒2t(t-1)+5(t-1)
⇒2t+5 and t-1 are the factors
therefore time can 1 or -5/2
as time cannot be negative we get time is 1
by this
h=1/2*10*1
h=5 m
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