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## Homework Statement

A cannon which is at the top of a 99 m high cliff fires in a horizontal direction to a dog. The initial position of the dog is 1000 m from the cliff's base. What should be the initial velocity of the cannon if the dog is moving at constant speed of 9 m/s to the cliff.

Other given

square root of (198 m/9.8 m/s) = 4.49 s

## Homework Equations

Yf = Yi + Vi(t) + (1/2)(gt^2)

Xf = Xi + Vi(t) + (1/2)(gt^2)

## The Attempt at a Solution

For the y-component:

99 m = 0 + Vi(4.49 s) + (1/2)(-9.8 m/s^2)(4.49)^2

= Vi(4.49 s) -99.78 m - 99 m

= Vi(4.49 s) - 198.78 m

For the x-component:

0 m = 1000 m + (-9 m/s)(4.49 s) + (1/2)(-9.8 m/s^2)(4.5)^2

= 860.365 m

Then,

Xi - Xf + Vi(t) + (1/2)(gt^2) = Yi - Yf + Vi(t) + (1/2)(gt^2)

Vi(4.49 s) - 198.78 m = 860.365 m

Vi(4.49 s) = 860.365 + 198.78 m

Vi = 1059.145 m/4.49 s

Vi = 235.89 m/s

Can someone verify this? Thanks alot.