Question

the equation of trajectory of a projectile thrown from a point on the ground is Y =x-x^2/40 m .g= 10 m/s^2 the maximum height reached is​

Answer 1

Answer:

according to eqn of trajectory ans is 10

Answer 2

Answer:

The maximum height reached is $10 \mathrm{~m}$.

Explanation:

Given:

$Y =x-x^2/40 m$ and

$g= 10 m/s^2$

To find the maximum height reached.

Step 1

The equation of trajectory of a projectile thrown with Initial speed u making an angle $\theta$ with horizontal is,

$y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}$

which can be rewritten as,

$y=x \tan \theta\left[1-\frac{x}{R}\right]$

Step 2

Where $R$ is the horizontal range,

For a projectile, It must be noted that,

$& \frac{\mathrm{H}}{\mathrm{R}}=\frac{\tan \theta}{4}$

$&\frac{1}{\mathrm{R}}=\frac{\tan \theta}{4 \mathrm{H}}$

Where $H$ is the maximum height attained by the projectile

$y=x \tan \theta\left[1-\frac{x \tan \theta}{4 \mathrm{H}}\right] \quad \ldots(2)$

Step 3

In the question, we're given,

$y=x-\frac{x^{2}}{40}$

$y=x\left(1-\frac{x}{40}\right)$

Comparing this equation with $(2)$ we get,

$\tan \theta=1$

and,

$\frac{\tan \theta}{4 \mathrm{H}}=\frac{1}{40}$

${H=10 \mathrm{~m}}$

Hence, the maximum height reached is $10 \mathrm{~m}$.

#SPJ2