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Assume that a projectile is fired horizontally from a "gun" located 1 meter above the ground. If the ball strikes the ground a distance of 2 meters from the end of the "gun" determine the muzzle velocity of the gun.

User Zash
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1 Answer

4 votes

Answer:


v_(o)=8.85m/s

Step-by-step explanation:

To determine the muzzle velocity of the gun, we must know how long does it take the ball to strikes the ground


y=y_(o)+v_(oy)t+(1)/(2)gt^(2)

Since the ground is at y=0 and
v_(oy)=0


0=1-(1)/(2)(9.8)t^(2)

Solving for t


t=0.4517s

Now, to determine the muzzle velocity we need to find its acceleration first


x=x_(o)+v_(ox)t+(1)/(2)at^(2) (1)


v=v_(ox)+at (2)

If we analyze the final velocity is 0. From (2) we have that


v_(ox)=-at (3)

Replacing (3) in (1)


2=-at^(2)+(1)/(2)at^(2)


2=a(0.4517)^(2) ((1)/(2)-1)


a=-19.60m/s^(2)

Solving (3)


v_(ox)=-at=-(19.60m/s^(2) )(0.4517s)=8.85m/s

User Benallansmith
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