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.

Question

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)

Henry · Answer

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.