Question

A projectile moves from the ground such that its horizontal displacement is $x=kt$ and vertical displacement is $y=kt(1-\alpha t)$ where k and $\alpha$ are constant and t is time. Find out total time of flight $\left(T\right)$ and maximum height attained $\left(Y_{max}\right).$

• A. $T=\alpha ,Y_{max}=\frac{k}{2a}$
• B. $T=\frac{1}{\alpha },Y_{max}=\frac{2k}{\alpha }$
• C. $T=\frac{1}{\alpha },Y_{max}=\frac{k}{6\alpha }$
• D. $T=\frac{1}{\alpha },Y_{max}=\frac{k}{4\alpha }$

Answer 1

The correct option is D $T=\frac{1}{\alpha },Y_{max}=\frac{k}{4\alpha }$

$\text{Step 1: Time of flight [Refer Fig.]}$

Here $x=kt$

$y=kt(1-\alpha t)$

For total time of flight $T$ displacement in y direction will be zero.

$\Rightarrow$ $y=0$

$\therefore 0=kt(1-\alpha t)$

$\Rightarrow t=0$ (Not possible)

or

$t=\frac{1}{\alpha }$

$\Rightarrow$ $T=\frac{1}{\alpha }$

$\text{Step 2: Maximum hight}{Y}_{\text{max}}$

At maximum hight point P velocity in y direction will be zero

$V_y=0$ and $\frac{dy}{dt}=V_y$

$\therefore \frac{dy}{dt}=0$

$\Rightarrow$ $\frac{d[kt-\alpha k{t}^2]}{dt}=0$

$\Rightarrow$ $k-2\alpha kt=0$

$\Rightarrow$ $t=\frac{1}{2\alpha }$

$Y=kt(1-\alpha t)$

$Y_{\text{max}}=k\left(\frac{1}{2\alpha }\right)[1-\frac{\alpha }{2\alpha }]$

$Y_{\text{max}}=\frac{k}{4\alpha }$

Hence option $D$ is correct.