Question 15: Work, Energy and Power & Circular Motion

Question

A metal ball of $$20 \mathrm{~g}$$ is projected at an angle $$30^{\circ}$$ with the horizontal with an initial velocity $$10 \mathrm{~ms}^{-1}$$. If the mass and angle of projection are doubled keeping the initial velocity the same, the ratio of the maximum height attained in the former to the latter case is :

Accepted Answer

To determine the ratio of the maximum heights attained in the two cases, let's start by deriving the formula for the maximum height reached by a projectile. The maximum height $$H$$ for a projectile is given by:
$$ H = \frac{u^2 \sin^2(	heta)}{2g} $$
Here:
$$u$$ is the initial velocity
$$	heta$$ is the angle of projection
$$g$$ is the acceleration due to gravity
For the first case:
Initial velocity, $$u = 10 	ext{ m/s}$$
Angle of projection, $$	heta = 30^\circ$$
Substituting these values into the formula:
$$ H_1 = \frac{(10)^2 \sin^2(30^\circ)}{2 	imes 9.8} $$
We know that $$ \sin(30^\circ) = \frac{1}{2} $$, so:
$$ H_1 = \frac{100 \left( \frac{1}{2} ight)^2}{2 	imes 9.8} = \frac{100 	imes \frac{1}{4}}{2 	imes 9.8} = \frac{25}{19.6} $$
For the second case, the mass and the angle of projection are doubled, but the mass does not affect the height. Therefore, the new angle of projection is:
$$ 	heta' = 2 	imes 30^\circ = 60^\circ $$
Using the same initial velocity $$u = 10 	ext{ m/s}$$, we substitute into the formula:
$$ H_2 = \frac{(10)^2 \sin^2(60^\circ)}{2 	imes 9.8} $$
We know that $$ \sin(60^\circ) = \frac{\sqrt{3}}{2} $$, so:
$$ H_2 = \frac{100 \left( \frac{\sqrt{3}}{2} ight)^2}{2 	imes 9.8} = \frac{100 	imes \frac{3}{4}}{2 	imes 9.8} = \frac{75}{19.6} $$
Now, to find the ratio of the maximum heights, we calculate:
$$ 	ext{Ratio} = \frac{H_1}{H_2} = \frac{\frac{25}{19.6}}{\frac{75}{19.6}} = \frac{25}{75} = \frac{1}{3} $$
Therefore, the ratio of the maximum height attained in the former to the latter case is 1 : 3, which corresponds to Option C.
Option C: 1 : 3

Answer

1 : 2

Answer

2 : 1

Answer

1 : 3

Answer

3 : 1

Work, Energy and Power & Circular Motion

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