A cannonball is catapulated toward a castle. The cannonball\'s velocitywhen it l

Question

A cannonball is catapulated toward a castle. The cannonball's velocitywhen it leaves the catapult is 40 m/s at an angle of 37 degrees with respect to the horizontal and the cannonball is 7.0 m above ground at this time.

a). What is the maximum height above the ground that was 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 land?

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

Explanation / Answer

Initial y component of launch = (sin 37) x 40 = 24.07m/sec.
x component = (cos 37) x 40 = 31.94m/sec.
y height gain = (v^2/2g) = (24.07^2/19.6) = 29.55 metres.
Time to max. height = (v/g) = 24.07/9.8 = 2.45 secs.
(29.55 + 7) = 36.55m. above ground.
a)Maximum height above the ground =36.55m
Time to drop 36.55m. = sqrt.(2h/g) = 2.72 secs.
Time in air = (2.72 + 2.45) = 5.17 secs.

b) Horizontal distance = (5.17 x 31.94) = 165.12 metres.
c)
Vertical V (y) component at ground = sqrt.(2gh) = sqrt.(19.6 x 36.55) = 26.76m/sec.
Horizontal V (x) component is 31.94m/sec.

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