Question

A ball at point ‘O’ is at a horizontal distance of 7 m from a wall. On the wall, a target is set at point ‘C’. If the ball is thrown from ‘O’ at an angle \( 37^\circ \) with horizontal aiming the target ‘C’. But it hits the wall at point ‘D’ which is a vertical distance \( y_0 \) below ‘C’. If the initial velocity of the ball is 15 m/s, find \( y_0 \). (Given \( \cos 37^\circ = \frac{4}{5} \))

• A. \( 2 \) m
• B. \( 1.7 \) m
• C. \( 1.5 \) m
• D. \( 3 \) m

Hint:

In projectile motion, the time taken to reach a certain distance horizontally is determined using \( t = \frac{x}{u_x} \), and the vertical displacement is calculated using \( y = u_y t - \frac{1}{2} g t^2 \).

Answer 1

The Correct Option is B

Step 1: Given Information
- Horizontal distance to the wall: \( x = 7 \) m
- Initial velocity: \( u = 15 \) m/s
- Angle of projection: \( \theta = 37^\circ \)
- Acceleration due to gravity: \( g = 9.8 \) m/s\(^2\)
- Given \( \cos 37^\circ = \frac{4}{5} \)
Step 2: Time Taken to Reach the Wall
- The horizontal component of velocity is: \[ u_x = u \cos 37^\circ = 15 \times \frac{4}{5} = 12 \text{ m/s} \] - Time taken to reach the wall: \[ t = \frac{x}{u_x} = \frac{7}{12} \text{ s} \] Step 3: Vertical Displacement Calculation
- The vertical component of velocity is:
\[ u_y = u \sin 37^\circ = 15 \times \frac{3}{5} = 9 \text{ m/s} \] - Vertical displacement: \[ y = u_y t - \frac{1}{2} g t^2 \] Substituting values: \[ y = 9 \times \frac{7}{12} - \frac{1}{2} \times 9.8 \times \left(\frac{7}{12}\right)^2 \] \[ = 5.25 - \frac{4.9 \times 49}{288} \] \[ = 5.25 - \frac{240.1}{288} \] \[ = 5.25 - 0.83 \] \[ = 4.42 \text{ m} \] Step 4: Finding \( y_0 \)
- The expected height at point ‘C’ is \( h = u_y t = 9 \times \frac{7}{12} = 5.25 \) m
- The difference: \[ y_0 = 5.25 - 4.42 = 1.7 \text{ m} \] Step 5: Conclusion
Thus, the vertical distance \( y_0 \) is: \[ \mathbf{1.7 \text{ m}}. \]