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 metre) respectively vary with time t in second as, x = (10√3) t and y = 10t -t^2. The maximum
height attained by the body is
(a) 200 m
(b) 100 m
(c) 50 m
(d) 25 m​

Answer 1

Given:

The horizontal and vertical displacements x and y(in metre) respectively vary with time t in second as, x = (10√3) t and y = 10t -t².

To find:

Maximum height reached.

Calculation:

First , let's find out the equation of trajectory of the object.

$\sf{x = 10 \sqrt{3} t \: \: and \: \: y = 10t - {t}^{2} }$

Putting value of t in 2nd Equation:

$\therefore \: y = 10t - {t}^{2}$

$= > \: y = 10( \dfrac{x}{10 \sqrt{3} } )- {( \dfrac{x}{10 \sqrt{3} } )}^{2}$

$\boxed{= > \: y = \dfrac{x}{ \sqrt{3} } - \dfrac{ {x}^{2} }{300}}$

Now, putting y = 0 in order to get range for the projectile motion:

$= > \: 0 = \dfrac{x}{ \sqrt{3} } - \dfrac{ {x}^{2} }{300}$

$= > \: \dfrac{x}{ \sqrt{3} } = \dfrac{ {x}^{2} }{300}$

$= > \: x = \dfrac{300}{ \sqrt{3} }$

$= > \: x = \dfrac{100 \times 3}{ \sqrt{3} }$

$= > \: x = 100 \sqrt{3}$

Now, maximum height will be obtained at half of the range.

So, putting value of x = 50√3 m in the Equation:

$= > \: y = \dfrac{x}{ \sqrt{3} } - \dfrac{ {x}^{2} }{300}$

$= > \: y = \dfrac{50 \sqrt{3} }{ \sqrt{3} } - \dfrac{ {(50 \sqrt{3}) }^{2} }{300}$

$= > \: y = 50 - 25$

$= > \: y = 25 \: m$

So, maximum height obtained is 25 metres.