Question

A cricketer hits a ball at $45^{\circ}$ with velocity $40\ \text{m s}^{-1}$ and it falls at $160\ \text{m}$. If he hits at the same angle with $50\ \text{m s}^{-1}$, the distance will be:

• A. 480 m
• B. 180 m
• C. 280 m
• D. 300 m
• E. 250 m

Hint:

For a fixed launch angle, doubling the velocity quadruples the range.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This problem deals with projectile motion. The distance the ball travels horizontally before hitting the ground is called the range. The range depends on the initial velocity and the angle of projection.
Step 2: Key Formula or Approach:
The formula for the horizontal range (R) of a projectile is given by: \[ R = \frac{u^2 \sin(2\theta)}{g} \] where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.
From this formula, we can see that if the angle \( \theta \) is constant, the range is directly proportional to the square of the initial velocity: \[ R \propto u^2 \] This means we can set up a ratio: \[ \frac{R_2}{R_1} = \frac{u_2^2}{u_1^2} \] Step 3: Detailed Explanation:
We are given two scenarios:
Scenario 1: - Angle, \( \theta_1 = 45^\circ \) - Initial velocity, \( u_1 = 40 \text{ m/s} \) - Range, \( R_1 = 160 \text{ m} \) Scenario 2: - Angle, \( \theta_2 = 45^\circ \) (same angle) - Initial velocity, \( u_2 = 50 \text{ m/s} \) - Range, \( R_2 = ? \) Since the angle is the same in both cases, we can use the proportionality relationship.
\[ \frac{R_2}{R_1} = \left(\frac{u_2}{u_1}\right)^2 \] Substitute the given values: \[ \frac{R_2}{160} = \left(\frac{50}{40}\right)^2 \] \[ \frac{R_2}{160} = \left(\frac{5}{4}\right)^2 = \frac{25}{16} \] Now, solve for \( R_2 \): \[ R_2 = 160 \times \frac{25}{16} \] \[ R_2 = 10 \times 25 = 250 \text{ m} \] Step 4: Final Answer:
The ball will fall at a distance of 250 m.