Answer:
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
Step-by-step 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.