John is fired horizontally from a cannon with the speed of 24.5 m/s. In 3.0 seconds he strikes the earth down range. A net will be located on the ground to catch him. a) what is his initial position above the ground? b) What is the horizontal range to put the net?

a)+ upward direction
x0= ?
xf=x0+v0cos(0)t

0= x0+ (24.5)cos(0)(3.0)
x0=73.5 m

b) x= ?
x=x0+v0cos(0)t

x=73.5+24.5cos(0)(3.0)
x=147m

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1. (b) The horizontal speed is unaffected by gravity, so it remains 24.5 m/s.

In 3 seconds, John travels 3*24.5 = 73.5 meters.

(a) we know that the vertical position, starting at y0 is
y(t) = y0 - 4.9t^2
Since y(3)=0, we can get the initial height by solving

y0 - 4.9*9 = 0
y0 = 44.1
Wow - that's a pretty high cannon!

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posted by Steve

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