Question

The maximum horizontal range of a ball projected from the ground is 32 m. If the ball is thrown with the same speed horizontally from the top of a tower of height 25 m, the maximum horizontal distance covered by the ball is: (Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)) \vspace{0.5cm}

• A. \( 40 \) m
• B. \( 57 \) m
• C. \( 60 \) m
• D. \( 75 \) m \vspace{0.5cm}

Hint:

For horizontal motion, the range remains the same when thrown from a height, and the total horizontal distance is given by \( x = ut \), where \( u \) is the horizontal velocity.

Answer 1

The Correct Option is A

Step 1: Understanding the Given Data - The maximum horizontal range \( R \) for projectile motion from the ground is given as 32 m. - The same initial horizontal speed is used when thrown from a height of 25 m. - We need to determine the horizontal distance covered before hitting the ground. \vspace{0.5cm} Step 2: Time of Flight from the Tower The time of flight is determined by the vertical motion: \[ h = \frac{1}{2} g t^2 \] Substituting values: \[ 25 = \frac{1}{2} \times 10 \times t^2 \] \[ t^2 = 5 \Rightarrow t = \sqrt{5} \approx 2.24 \, \text{seconds} \] \vspace{0.5cm} Step 3: Horizontal Distance Covered The horizontal speed \( u \) is the same as in the projectile motion equation: \[ R = \frac{u^2}{g} \Rightarrow 32 = \frac{u^2}{10} \] \[ u^2 = 320 \Rightarrow u = \sqrt{320} \approx 17.89 \, \text{m/s} \] The horizontal distance covered in time \( t \) is: \[ \text{distance} = u \times t = 17.89 \times 2.24 \] \[ = 40 \text{ m} \] Thus, the maximum horizontal distance covered is: \[ \mathbf{40} \text{ m} \] \vspace{0.5cm}