Posted by **Anonymous** on Thursday, February 16, 2012 at 3:51pm.

A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 39 m/s at an angle of 40° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time.

(a) What is the maximum height above the ground reached by the cannonball?

1 m

(b) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land?

2 m

(c) What are the x- and y-components of the cannonball's velocity just before it lands? The y-axis points up.

3 m/s (x-component)

4 m/s (y-component)

- physics3 -
**Henry**, Saturday, February 18, 2012 at 11:22am
Vo = 39m/s @ 40 Deg.

Xo = 39*cos40 = 29.9 m/s.

Yo = 39*sin40 = 25.1 m/s.

a. hmax = ho + (Yf^2-Yo^2)/2g,

hmax = 7 + (0-(25.1)^2) / -19.8=39.1 m.

b. Dx = Vo^2*sin(2A)/g, 2A = 80 Deg.

c. Xo = 39*cos40 = 29.9 m/s.

Yf^2 = Vo^2 + 2g*hmax,

Yf^2 = 0 + 19.6*39.1 = 766.36,

Yf = 27.7 m/s.

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