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Maslowich
3 years ago
9

Kyle is flying a helicopter and is rising at 5.0m/s when he releases the bag. After 2.0s

Physics
2 answers:
ELEN [110]3 years ago
8 0
A. What is the bag’s velocity?
Answer: 25 m/s 

b. How far has the bag fallen?
Answer: -30 m

c. How far below the helicopter is the bag?
Answer: 20 m below the helicopter

ch4aika [34]3 years ago
5 0
1) v=u+gt = -5 +10*2=15 m/s
2) r=ut+gt^2/2=-5*2+10*2^2/2=-10+20=10m
3) r of heli= vt =5*2=10m
So 10m by bag from answer number 2 plus additional 10m by helicopter equals 20m
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3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
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Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

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x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

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t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

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Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

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x=14.49 m

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