Question

A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 670 m/s. The gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.025 m below the center. a What is the horizontal distance between the end of the rifle and the bull's-eye? b. If the target were not in the way, how far would the bullet travel horizontally before hitting the ground? PLEASE INCLUDE ALL UNITS AND CONVERSIONS IN CALCULATIONS

Answer 1

a).
`x=47.8m`

b).
`x=246.12m`

Step-by-step explanation:

a).

`x=v_(ox)*t`

`y=y_i+v_(oy)+(1)/(2)*a_y*t^2`

`v_(ox)=670 m/s`

`y=(1)/(2)*a_y*t^2`

`y=0.025m`

`t=\sqrt{(2*-y)/(-g)}=\sqrt{(-2*0.025m)/(-9.8m/s^2) }`

`t=0.0714s`

`x=v_(ox)*0.0714s`

`x=670*0.0714=47.8m`

b).

Assume a normal person fired the rifle with a high of 1.8m so:

`x=v_(ox)*\sqrt{(2*h)/(g)}`

`x=670(m/s)*\sqrt{(2*1.8m)/(9.8m/s^2)}`

`x=246.12m`