Question

A ball of mass 522 g starts at rest and slides down a frictionless track. It leaves the track horizontally, striking the ground a distance of 1 m from the end of the track after falling a vertical distance 1.25 m from the end of the track. (

Answer 1

Force on the ball after falling  a vertical distance is 5.1156 N and the final velocity at the ground is 4.949 m/s.

Explanation:

Given:

mass of the ball = 522 g = 0.522 kg

Track distance = 1 m

 Vertical distance = 1.25 m

 Gravitational acceleration = 9.8 $m/s^{2}$

Initial velocity = 0 m/s

F = mg

  = 0.522 x 9.8

  = 5.1156 N

    ⇒ $v^{2} - u^{2} = 2gs$

        ⇒ v = $\sqrt{2 \times 9.8\times 1.25 }$

                 = 4.949 m/s