Ask Question
15 June, 00:53

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?

+5
Answers (1)
  1. 15 June, 01:58
    0
    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.
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers