Question

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.

Answer 1

`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`