In a spring gun having spring constant 100 N/m a small ball 'B' of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground. The value of d is _________ m. (g = 10 m/s²).

Solution

Physics sir · 3 steps

The spring's stored energy gives the launch speed, and the 2 m fall gives the flight time.

1. Spring to speed. All the elastic energy becomes the ball's kinetic energy: 1/2kx² = 1/2mv². So v² = kx²/m = (100 × (0.05)²)/0.1 = 2.5 m² s⁻², which gives v = √2.5 ≈ 1.58 m s⁻¹, horizontal.
2. Time of fall. The ball leaves horizontally, so its vertical speed starts at zero. From h = 1/2gt², the time is t = √(2h/g) = √((2 × 2)/10) = √0.4 ≈ 0.63 s.

   d = vt = √2.5 × √0.4 = √1 = 1 m

3. So. The horizontal speed stays constant during the fall, so the box must be placed at d = 1 m.

The idea to keep. For a horizontal launch, the height alone fixes the flight time through t = √(2h/g). Find the launch speed from energy, then range is speed times time.

Answer from the NTA final answer key. Solution by PrepWiser.