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

Answer:

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

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.