## 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 negl

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 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.

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3 years 2021-08-30T13:38:45+00:00 1 Answers 36 views 0

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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.