Question 56: Motion in 2D & Vectors

Question

The trajectory of projectile, projected from the ground is given by $y=x-\frac{x^{2}}{20}$. Where $x$ and $y$ are measured in meter. The maximum height attained by the projectile will be.

Accepted Answer

The equation of the trajectory given is $y = x - \frac{x^2}{20}$.
This is a parabola, and it represents the path of the projectile.
The maximum height of the projectile corresponds to the vertex of the parabola.
The x-coordinate of the vertex for a parabola given by $y = ax^2 + bx + c$ is $-b/2a$.
In this case, $a = -1/20$ and $b = 1$, so the x-coordinate of the vertex is:
$ x_{	ext{vertex}} = -\frac{b}{2a} = -\frac{1}{2 	imes (-1/20)} = 10 $
Substituting this into the equation of the trajectory gives the y-coordinate of the vertex, which is the maximum height:
$ y_{	ext{max}} = 10 - \frac{10^2}{20} = 10 - 5 = 5 $
So, the maximum height attained by the projectile is 5 m.

Answer

10 m

Answer

5 m

Answer

200 m

Answer

10$\sqrt2$ m

Motion in 2D & Vectors

The trajectory of projectile, projected from the ground is given by y=x-\frac{x^{2}}{20}. Where x and y are measured in meter. The maximum height attained by the projectile will be.

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