Question

When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed should you throw it up vertically so it will go twice as high?

and

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

The answer of first question:

The speed should be √2 times the initial speed.

v°=√2×vi

v° stands for new speed

vi stands for initial speed

The answer of second question:

The height should be four times the previous height.

h°=4×h

h° stands for new height

h stands for previous height

Step-by-step explanation:

According to conservation of energy:

The energy can't be created or destroyed but it can transform from one form to another form of energy.

When a pebble throw upward it's kinetic energy converted into potential energy. So, according to law of conservation of energy

Kinetic energy = Potential energy

( 1/2 )m×v² = m×g×h

m stands for mass of pebble

v stands for velocity

g stands for gravitational constant

h stands for height

(1/2) m×v² = m×g×h

Multiply by 2 and divide by m on both sides

v² = 2×g×h

Take square root on both sides

v =√(2×g×h) ...................Equation (A).

If we want twice the height (h°=2×h), then the new velocity will be v°

Again using law of conservation of energy

Kinetic energy = Potential energy

(1/2) m×v°² = m×g×h°

v° stands for new velocity

h° stands for new height

put h°=2×h in above equation

(1/2) m×v°² = m×g×2×h

Multiply by 2 and divide by m on both sides

v°² = 2×g×2×h

Take square root on both sides

v° = √(2×g×2×h)

v° = √2×√(2×g×h)

from equation (A) we know that v=√(2×g×h)

So the equation become

v°= √2×v

The speed should be √2 times the initial speed.

Explanation for second answer

According to law of conservation of energy

Kinetic energy = Potential energy

(1/2)× m×v² = m×g×h

Divide by m on both sides

(1/2) × v² = g×h

Divide by g on both sides

v²/(2×g)= h .............Equation (B)

If we want twice the speed (v°=2×v), then the new height will be h°

According to law of conservation of energy

Kinetic energy = Potential energy

(1/2) m×v°² = m×g×h°

v° stands for new velocity

h° stands for new height

(1/2) m×v°² = m×g×h°

put v° = 2×v in above equation

(1/2) m×(2×v)²=m×g×h°

(1/2) m× 4×v²=m×g×h°

Divide by m on both sides

(1/2) 4×v²=g×h°

Divide by g on both sides

4 (v²/2×g) =h°

From equation (B) we know that v²/(2×g)

So the equation become

h° = 4×h

The height should be four times the previous height.