From the top of a tower a stone is thrown up which reaches the ground in a time `t_(1)`. A second stone thrown down with same speed reaches the ground in a time `t_(2)`. A third stone released from rest from the same location reaches the gound in a time `t_(3)`. then

Options

A. `(1)/(t_(3))=(1)/(t_(2))-(1)/(t_(1))`
B. `t_(3)^(2)=t_(1)^(2)-t_(2)^(2)`
C. `t_(3)=(t_(1)+t_(2))/(2)`
D. `t_(3)=sqrt(t_(1)t_(2))`

Answer

D

Explanation

Let h be the height of the tower. Then from the laws of motion (under gravity), we get
`h=-ut_(1)+(1)/(2)g t_(1)^(2)` . . (i), When projected upwards with velocity u.
`h=ut_(2)+(1)/(2)g t_(2)^(2)` . . (ii) , When thrown downwards with velocity u.
Also, `h=(1)/(2)g t_(3)^(2)` . . (iii) , when releases from rest.
From (i), we get `(h)/(t_(1))=-u+(1)/(2)g t_(1)` . . (iv)
From (ii), we get `(h)/(t_(2))=u+(1)/(2)g t_(2)` . . (v)
Adding (iv) and (v), we get
`h((1)/(t_(1))+(1)/(t_(2)))=(1)/(2)g(t_(1)+t_(2))`,
putting the values of h from (iii)
`impliest_(3)=sqrt(t_(1)t_(2))`.