A ball is thrown upwards at an angle of `60^(@)` to the horizontal. It falls on the ground at a distance of 90 m . If the ball is thrown with the same initial velocity at an angle `30^(@)`, it will fall on the ground at a distance of

Options

A. `30m`
B. `60 m`
C. `90 m`
D. `120 m`

Answer

To solve the problem, we can use the concept of projectile motion. The horizontal range of a projectile is given by the formula:

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

where:
- \( R \) = range of the projectile,
- \( u \) = initial velocity,
- \( \theta \) = angle of projection,
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 1: Calculate the initial velocity using the first angle (60 degrees)

Given that the range \( R \) for the angle \( \theta = 60^\circ \) is 90 m, we can set up the equation:

\[ 90 = \frac{u^2 \sin(2 \times 60^\circ)}{g} \]

Calculating \( \sin(120^\circ) \):

\[ \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

Now substituting this into the equation:

\[ 90 = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]

Rearranging gives:

\[ u^2 = \frac{90g}{\frac{\sqrt{3}}{2}} \]

\[ u^2 = \frac{180g}{\sqrt{3}} \]

### Step 2: Calculate the range for the second angle (30 degrees)

Now, we need to find the range when the ball is thrown at an angle of \( 30^\circ \):

Using the same formula:

\[ R' = \frac{u^2 \sin(2 \times 30^\circ)}{g} \]

Calculating \( \sin(60^\circ) \):

\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

Now substituting \( u^2 \) into the range formula:

\[ R' = \frac{\left(\frac{180g}{\sqrt{3}}\right) \cdot \frac{\sqrt{3}}{2}}{g} \]

The \( g \) cancels out:

\[ R' = \frac{180 \cdot \frac{\sqrt{3}}{2}}{\sqrt{3}} \]

Simplifying gives:

\[ R' = \frac{180}{2} = 90 \, \text{m} \]

### Final Answer

Thus, when the ball is thrown at an angle of \( 30^\circ \), it will fall on the ground at a distance of **90 m**.

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