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 $ .