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

Answer 1

Hint : At the highest point, the velocity of the particle will be zero and the displacement will have the maximum value, i.e. $H$ . We will use the formulas of projectile motion in a plane to get the required equations.

Formula used:
$\Rightarrow v = u + at$
where $v$ is the final velocity of the particle, $u$ is the initial velocity of the particle, $a$ is the acceleration acting on the particle, and $t$ is the time of action and for the calculation of the final velocity.
$\Rightarrow S = ut + \dfrac{1}{2}a{t^2}$ ,
where $S$ is the displacement of the body and the rest of the notations are the same as in the above equation.

Complete step by step answer
Applying the formula $v = u + at$ at the topmost point, we get
$\Rightarrow 0 = U + ( - g)T$
$\Rightarrow U = gT$ ,
where $U$ is the initial velocity of the particle.
Applying the formula $S = ut + \dfrac{1}{2}a{t^2}$ at the topmost point again, we get
$\Rightarrow H = UT + \dfrac{1}{2}( - g){T^2}$
Putting the value of the initial velocity $U$ as found above, we get
$\Rightarrow H = (gT)T - \dfrac{1}{2}g{T^2}$
$\Rightarrow H = \dfrac{1}{2}g{T^2}$ .
We will use these values in the general equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ , so that at a general height $h$ , we get the required equation as
$\Rightarrow h = Ut + \dfrac{1}{2}( - g){t^2}$ ,
substituting $U = gT$ , we get
$\Rightarrow h = gTt - \dfrac{1}{2}g{t^2}$
Adding $0 = H - \dfrac{1}{2}g{T^2}$ to the above equation we get,
$\Rightarrow h = gTt - \dfrac{1}{2}g{t^2} + H - \dfrac{1}{2}g{T^2}$
$\Rightarrow h = H + (gTt - \dfrac{1}{2}g{t^2} - \dfrac{1}{2}g{T^2})$
On rearranging the equation further, we get,
$\Rightarrow h = H - \dfrac{1}{2}g( - 2Tt + {t^2} + {T^2})$
$\Rightarrow h = H - \dfrac{1}{2}g{(t - T)^2}$
Therefore, the correct answer is option (B); $H - \dfrac{1}{2}g{(t - T)^2}$ .

Note
The answer obtained is symmetric for a time of $T$ before and after the value of $t = T$ . This means that the value of the height of the particle will be the same at $t = 0$ and at $t = 2T$ . This indicated that the displacement of the particle will start from $0$ , reach the highest value of $H$ at $t = T$ , and then decrease, at the same rate of increase, to $0$ at $t = 2T$ .