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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. $g{{\left( t-T \right)}^{2}}$
B.$H-\dfrac{1}{2}g{{\left( t-T \right)}^{2}}$
C.$\dfrac{1}{2}g{{\left( t-T \right)}^{2}}$
D. $H-g{{\left( t-T \right)}^{2}}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: Firstly, you could use the equations of motion for the body at maximum height H and time T. Then you could make use of the equation of motion for some height h at time t. After that, you could relate the two equations and then make necessary changes and thus find the required relation.

Formulae used:
Equations of motion,
$v=u+at$
$s=ut+\dfrac{1}{2}g{{t}^{2}}$

Complete Step by step solution:
In the question, we are given a particle that is being projected vertically upwards. We are given that the particle reaches a maximum height H in time T seconds and we are asked to find the height of the particle at any given time t.

We know that at maximum height the velocity of the particle will be zero, that is,
$v=0m{{s}^{-1}}$
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Let the initial velocity of the particle be u when projected, then, by equation of motion we have,
$v=u+at$
$\Rightarrow 0=u-gT$
$\therefore u=gT$ ………………………………. (1)

Also, we have another equation of motion given by,
$H=uT-\dfrac{g{{T}^{2}}}{2}$

Substituting (1), we get,
$\Rightarrow H=g{{T}^{2}}-\dfrac{g{{T}^{2}}}{2}$
$\therefore H=\dfrac{g{{T}^{2}}}{2}$ ………………………………………….. (2)

Let the height of the particle be h for any time t, then, by equation of motion we have,
$s=ut+\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow h=ut-\dfrac{1}{2}g{{t}^{2}}$
Substituting (1),
$h=gTt-\dfrac{1}{2}g{{t}^{2}}$

Now let us add and subtract $\dfrac{1}{2}g{{T}^{2}}$ on the right hand side, then,
$h=gTt-\dfrac{1}{2}g{{t}^{2}}-\dfrac{1}{2}g{{T}^{2}}+\dfrac{1}{2}g{{T}^{2}}$
$\Rightarrow h=\left( gTt-\dfrac{1}{2}g{{t}^{2}}-\dfrac{1}{2}g{{T}^{2}} \right)+\dfrac{1}{2}g{{T}^{2}}$

But from (2),$H=\dfrac{g{{T}^{2}}}{2}$ so, h becomes,
$h=-\dfrac{1}{2}g\left( -\dfrac{1}{2}Tt+{{t}^{2}}+{{T}^{2}} \right)+H$
$\therefore h=H-\dfrac{1}{2}g{{\left( t-T \right)}^{2}}$

Therefore, we found that the height of the particle at any time t will be,
$h=H-\dfrac{1}{2}g{{\left( t-T \right)}^{2}}$

Hence, option B is found to be the correct answer.

Note:
We know that, by convention we take acceleration due to gravity positive when the motion of the body is downwards. Since, here the body’s motion is upwards, we have taken it to be negative. Also, when a particle is projected vertically upwards, we know that it goes to rest and maximum height and returns downwards and hence the velocity at maximum height is zero.