Question

The equation of trajectory of a projectile thrown from a point on the ground is y=(x−x2/40)m. If g=10ms−2. The maximum height reached is

Answer 1

The equation of trajectory of a projectile thrown with initial speed $\sf{u}$ making an angle $\theta$ with horizontal is,

$\sf{\longrightarrow y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}}$

which can be rewritten as,

$\sf{\longrightarrow y=x\tan\theta\left[1-\dfrac{x}{R}\right]\quad\quad\dots(1)}$

where $\sf{R}$ is the horizontal range.

For a projectile it must be noted that,

$\sf{\longrightarrow\dfrac{H}{R}=\dfrac{\tan\theta}{4}}$

$\sf{\longrightarrow\dfrac{1}{R}=\dfrac{\tan\theta}{4H}}$

where $\sf{H}$ is the maximum height attained by the projectile.

$\sf{\longrightarrow y=x\tan\theta\left[1-\dfrac{x\tan\theta}{4H}\right]\quad\quad\dots(2)}$

In the question, we're given,

$\sf{\longrightarrow y=x-\dfrac{x^2}{40}}$

$\sf{\longrightarrow y=x\left(1-\dfrac{x}{40}\right)}$

Comparing this equation with (2) we get,

$\sf{\longrightarrow\tan\theta=1}$

and,

$\sf{\longrightarrow\dfrac{\tan\theta}{4H}=\dfrac{1}{40}}$

$\sf{\longrightarrow\underline{\underline{H=10\ m}}}$

Hence the maximum height reached is 10 m.