Question

A helicopter flying horizontally with a velocity of 288 kmph drops a bomb. If the line joining the point of dropping the bomb, and the point where bomb hits the ground makes an angle \( 45^\circ \) with the horizontal, then the height at which the bomb was dropped is (Acceleration due to gravity = \( 10 \, \text{ms}^{-2} \))

• A. 1320 m
• B. 1280 m
• C. 320 m
• D. 640 m

Hint:

If the line of sight angle is \( 45^\circ \) for a horizontal projectile, then Range = Height.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

The bomb undergoes projectile motion. It has a horizontal velocity \( u_x \) and zero initial vertical velocity \( u_y \). The line joining the drop point and impact point makes \( 45^\circ \) with the horizontal. This implies the horizontal range \( R \) is equal to the vertical height \( H \) (since \( \tan 45^\circ = \frac{H}{R} = 1 \)).
Step 2: Key Formula or Approach:

1. Horizontal Distance: \( R = u_x \times t \). 2. Vertical Distance: \( H = \frac{1}{2} g t^2 \). 3. Condition: \( H = R \).
Step 3: Detailed Explanation:

Convert velocity: \[ u_x = 288 \, \text{kmph} = 288 \times \frac{5}{18} \, \text{m/s} \] \[ u_x = 16 \times 5 = 80 \, \text{m/s} \] Since \( H = R \): \[ \frac{1}{2} g t^2 = u_x t \] Discarding \( t=0 \): \[ \frac{1}{2} g t = u_x \] \[ t = \frac{2 u_x}{g} \] Substitute \( t \) back into the equation for \( H \): \[ H = u_x \left( \frac{2 u_x}{g} \right) = \frac{2 u_x^2}{g} \] Calculate H: \[ H = \frac{2 \times (80)^2}{10} = \frac{2 \times 6400}{10} = 1280 \, \text{m} \]
Step 4: Final Answer:

The height is 1280 m.