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A cannon tilted up at a 26 degree angle fires a cannon ball at 71m/s from atop a 12m -high fortress wall.

What is the ball's impact speed on the ground below?

Added Wed, 08 Jul '15

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Total E = KEi + PEi

TotalE = KEf +PEf

PE = mgh

KE = 1/2mv^2

So Total E is always kept constant. it never changes. Thus it's equal to the KE + PE. Well, the PE of the cannon and thus the ball initially is mg(12). And the initial KE will be 1/2m(71)^2 **KE is a scalar quantity here**

Thus, the Total E of the ball will be 1/2m(71)^2 + m(9.8)(12)

That Total E is kept constant from start to finish (well, ignoring air resistance and whatnot). So thus at the end of its trip when the PE is equal to zero, the total E will be equal to the final KE

Total E = PEf + KEf

TotalE = 0 + 1/2mvf^2

Total E = TotalE

1/2m(71)^2 + m(9.8)(12) = 1/2mvf^2

**now you can see why the mass of the ball is neglected**

1/2(71)^2 + 117.6 = 1/2vf^2

71^2 + 2*117.6 = vf^2

vf^2 = 5276.2

vf = 72.637

Added Wed, 08 Jul '15

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