Posted by Anonymous on .
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?
(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?
(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)
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.