Question

A 0.050 kg ball starts from rest at some unknown height on a toy roller coaster.
At a later time, it travels through the top of a loop at 2 m/s and a height of 0.40 m.
Since this track is frictionless, what was the starting height of the ball?

Answer 1

Answer:

The starting height of the ball is approximately 0.604 m

Explanation:

The given parameters are;

The mass of the the ball = 0.050

The speed with which it travels through the top loop = 2 m/s

The given height at which the ball moves at 2 m/s = 0.40 m

Therefore, we have;

1/2·m·v² = m·g·h

1/2·v² = g·h

h = 1/2·v²/g = 1/2 × 2²/9.81 ≈ 0.204

The additional height = h = 0.204 m

Therefore;

The starting height of the ball  ≈ The given height at which the ball moves at 2 m/s + h

The starting height of the ball  ≈ 0.40 + 0.204 = 0.604 m

The starting height of the ball ≈ 0.604 m.

Answer 2

When any object in rest, then potential energy is present. But when object is in motion then object have kinetic energy.

Starting height of the ball is 0.604 m.

We know that, when any object is start from rest, then potential energy is converted into kinetic energy.

     $\frac{1}{2}mv^{2} =mgh$

Where m is mass of object, g is gravitational acceleration , h is height and v is velocity of object. (value of g = 9.81 m/ second square)

from above equation,

we get,     extra height        $h=\frac{v^{2} }{2g} \\\\h=\frac{4}{2*9.81}\\\\h=0.204$ meter

The starting height of the ball will be sum of the height at which ball moves 2 m/s and extra height.

Starting height = 0.40 + 0.204 = 0.604 meter.

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