A stone drop from height h reaches at earth surface in 1sec. If same s...
Answer:
Explanation of the scenario:
A stone is dropped from a height h on the earth's surface and it reaches the surface in 1 second. The time taken by the same stone to reach the surface of the moon when dropped from the same height is to be calculated.
Assumptions:
- The height h is assumed to be the same on both the earth and the moon.
- The gravitational force acting on the stone is assumed to be the same on both the earth and the moon.
Formula:
The formula to calculate the time taken by an object to fall freely from a certain height is given by:
t = √(2h/g)
where,
t = time taken
h = height
g = gravitational acceleration
Calculation:
On earth,
t = √(2h/g)
Given, t = 1 second
Squaring both sides, we get:
1 = 2h/g
Solving for h, we get:
h = g/2
On the moon, the gravitational acceleration is one-sixth of the earth's gravitational acceleration. Therefore,
gmoon = 1/6 * gearth
Substituting the value of gmoon in the above equation, we get:
hmoon = gmoon/2
hmoon = 1/12 * gearth
Final Answer:
Therefore, the time taken by the stone to reach the surface of the moon is calculated as follows:
tmoon = √(2hmoon/gmoon)
tmoon = √(2 * 1/12 * gearth / 1/6 * gearth)
tmoon = √(1/4)
tmoon = 1/2 seconds
Therefore, the stone will take 1/2 seconds to reach the surface of the moon when dropped from the same height as it was dropped on earth.