Question

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

• A. $g(t-T{)}^2$
• B. $H-\frac{1}{2}g(t-T{)}^2$
• C. $\frac{1}{2}g(t-T{)}^2$
• D. $H-g(t-T{)}^2$

Answer 1

The correct option is B $H-\frac{1}{2}g(t-T{)}^2$

At maximum height, $v=0$

Thus, from equations of motion:

$0=u-gT⟹u=gT$

Also,

$H=uT-\frac{1}{2}g{T}^2=g{T}^2-\frac{1}{2}g{T}^2$

$H=\frac{1}{2}g{T}^2$

Here, $H$ is the maximum height and $T$ is the time taken to reach its maximum height

At any time $t$,

$h=ut-0.5g{t}^2=gTt-0.5g{t}^2$

Add and subtract $0.5g{T}^2$ , we get,

$h=(gtT-0.5g{t}^2-0.5g{T}^2)+0.5g{T}^2$

For above putting $H=0.5g{T}^2$ we get

$h=H-\frac{1}{2}g(t-T{)}^2$

Hence option $\left(B\right)$ is correct