# A body is thrown vertically upwards from A. The top of a tower . It reaches the ground in time t_(1). It it is thrown vertically downwards from A

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A body is thrown vertically upwards from A. The top of a tower . It reaches the ground in time t_(1). It it is thrown vertically downwards from A with the same speed it reaches the ground in time t_(2), If it is allowed to fall freely from A. then the time it takes to reach the ground.
.
A. t=(t_(1)+t_(2))/(2)
B. t=(t_(1)-t_(2))/(2)
C. t=sqrt(t_(1)t_(2))
D. t=sqrt((t_(1))/(t_(2))

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Suppose the body be projected vertically upwards from A with a speed u_(0)
Using equation s=ut+((1)/(2))at^(92), we get
For first case: -h=u_(0)t_(1)-((1)/(2))g t_(1)^(2) ..(i)
For second case : -h=-u_(0)t_(2)-((1)/(2))g t_(2)^(2) ..(ii)
(i)-(ii)implies0=u_(0)(t_(2)+t_(1))t_(2)-((1)/(2))g t_(2)^(2)
impliesh=(1)/(2)g t_(1)t_(2) ..(iv)
For third case: u=0,t=?
-h=0xxt-((1)/(2))g t^(2) or h=((1)/(2))g t^(2) ..(v)
Combining Eq. (iv) and (v) we get
(1)/(2)g t^(2)=(1)/(2)g t_(1)t_(2) or t=sqrt(t_(1)t_(2))`