Question

A body is projected at time (t = 0) from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in meter) respectively vary with time t in seconds as, $x=10\sqrt{3}t$ and $y=10t-t^2$. Then, the maximum height attained by the body is :

• A. 200 m
• B. 100 m
• C. 50 m
• D. 25 m

Answer 1

The correct option is D 25 m

Given: $y=10t-t^2$
and $x=\left(10\sqrt{3}\right)t$
$y_{max}=?$
$\frac{d\left(y\right)}{dt}=10-2t=0\Rightarrow 10=2t$
$t=5\text{sec}$
Putting the value of t in $y,y=10\left(5\right)-5\times 5$
$=50-25=25m$

Aliter,
$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

Comparing with
$y=10t-t^2$
$u_y=10\text{and}{a}_y=-2$

$H=\frac{u_y^2}{2a_y}=\frac{100}{4}=25m$