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Q.
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
Motion in a Straight Line
Solution:
Using, $v=u+a t$, we get
$0=u-g T$ or $u=g T$
Further, $v^{2}=u^{2}+2 a s$ or $0=u^{2}-2 g H$
or $H=\frac{u^{2}}{2 g}=\frac{g^{2} T^{2}}{2 g}=\frac{g T^{2}}{2}$
Let $h$ be the distance travelled in time $t$. Then
$h=u t-\frac{1}{2} g t^{2}=g T \times t-\frac{1}{2} g t^{2}$
Subtracting (i) from (ii), we get
$h-H=g t T-\frac{1}{2} g t^{2}-\frac{g T^{2}}{2}=-\frac{g}{2}(t-T)^{2}$
$\therefore h=H-\frac{g}{2}(t-T)^{2}$