Question

When you drop a pebble from height H, it reaches the ground with speed V if there is no air resistance. From what height should you drop it so it will reach the ground with twice speed?

Answer 1

To reach the ground at twice the speed, the pebble should be dropped from a height 4 times the original height.

Step-by-step explanation:

To determine the height from which the pebble should be dropped so it will reach the ground at twice the speed, we can use the principle of conservation of energy.

When the pebble is dropped, it has potential energy due to its height that converts into kinetic energy when it reaches the ground. According to the principle of conservation of energy, the potential energy at height H is equal to the kinetic energy at the ground:

mg = 1/2mv^2

Where m is the mass of the pebble, g is the acceleration due to gravity, h is the height from which it was dropped, and v is the speed at which it reaches the ground.

If we want the pebble to reach the ground with twice the speed, we can substitute 2v for v in the equation:

mg = 1/2m(2v)^2

Simplifying the equation, we get:

h = 4H

So, to reach the ground at twice the speed, the pebble should be dropped from a height 4 times the original height.

Answer 2

H quadrupled

Step-by-step explanation:

Since we are neglecting air resistant, gravitational acceleration g is the only thing that affect the pebble:

As v = gt. To get twice the speed 2v with the same acceleration g, the time must be doubled.

Yet to get to distance H with 2t time. Since
`H = gt^2/2`. If t is doubled, H must be
`2^2 = 4` times the original amount.