A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be-

Options

A. `g(t-T)^(2)`
B. `H-1/2g (t-T)^2`
C. `1/2 g(t -T)^(2)`
D. `H - g (t - T)`

Answer

B

Explanation

Using ,v = u + at , we get
0 = u - gT or u = gT
Further , `v^2 =u^2 +2as " or " 0 = u^(2) - 2 gH`
or `H = u^2/(2g)=(g^2T^2)/(2g)=(gT^2)/2 " "...(i)`
Let h be the distance travelled in time t , Then .
`h = ut -1/2"gt"^(2) = "gT" xx t -1/2 "gt"^2 " "(ii)`
Subtracting (i) from (ii), we get
`h-H="gt" -1/2"gt"^2-(gT^2)/2=-g/2(t-T)^2`
`:. h = H -g/2 (t-T)^2`