Ask Question
10 February, 02:55

You throw a ball of mass 1.0 kg straight up. You observe that it takes 3.4 s to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up to the top is half the total time, 1.7 s. Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward.

+5
Answers (1)
  1. 10 February, 03:12
    0
    Incomplete question

    This is the complete question

    You throw a ball of mass 1 kg straight up. You observe that it takes 3.4s to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up to the top is half the total time, 1.7s. Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward.

    (a) Use the momentum principle to determine the speed that the ball had just AFTER it left your hand.

    vinitial = ? m/s

    (b) Use the Energy Principle to determine the maximum height above your hand reached by the ball.

    h = ? m

    Explanation:

    Mass of object is 1kg

    Total time to and fro is 3.4s

    i. e the time to reach maximum height will be half of total time = 1.7s

    g=9.81m/s²

    Question 1

    We need the initial velocity

    Using the free fall body

    v=u-gt. Against gravity upward

    at maximum height the body comes to rest before returning back, then v=0 at maximum height

    The time to reach max height is 1.7s

    So, v=u-gt

    0=u-9.81*1.7

    0=u-16.68

    Therefore, u=16.68m/s

    The initial velocity after the ball left the hand is 16.68m/s

    Now, using the principle of momentum

    The weight of the object is

    W=mg=9.81*1

    W=-9.81N. Downward

    Then, the force that is needed to throw the body upward is equal to the weight

    ΣF = mg=W

    F=W=-9.81

    From momentum

    F = (mv-mu) / t

    Since at max height, v=0

    Ft=mv-mu

    -9.81*1.7=0-1*u

    -16.677=-u

    Therefore, u=16.68m/s as expected from the equation of motion

    Question 2.

    Maximum height reach by the ball

    Using equation of motion

    v²=u²-2gH

    0²=16.68²-2*9.81*H

    0=278.12-19.62H

    19.62H=278.12

    Then, H=278.12/19.62

    H=14.18m

    Now using, energy principle

    Energy is conserved

    i. e kinetic energy is equal to P. E energy when the body is free falling

    K. E = ½mv²

    P. E=mgh

    K. E=P. E

    ½mv²=mgh

    ½v²=gh

    v²=2gh

    h=v²/2g

    Then,

    h=16.68²/2*9.81

    h=14.18m

    As expected, same as the equation of motion ... So both momentum and conservation of energy can be apply to free falling body.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “You throw a ball of mass 1.0 kg straight up. You observe that it takes 3.4 s to go up and down, returning to your hand. Assuming we can ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers