A cannonball is catapulted toward a castle. The cannonball\'s velocity when it l

Question

A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 48 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)

Explanation / Answer

Given : (a)cannonball's velocity when it leaves the catapult is = U = 48 m/s at an angle of = = 40° (with respect to the horizontal) cannonball is = h = 7.0 m ( above the ground at this time) (a) maximum Height above the ground reached by the cannon ball is Hmax = h + U2 sin2 / 2 g           = 7.0 + (48)2 sin2 (40) / 2 (9.8)           = 55.56 m (b) cannonball makes it over the castle walls and lands back down on the ground, at a horizontal distance from its release point will it land is = R = U2 sin2/ g                                = (48)2 sin2 (40) / 9.8                                = 231.53 m (c) x- and y-components of the cannonball's velocity just before it lands ( The y-axis points up)                            U cos40= 36.77m/s (x-component) U sin40= 30.85m/s   (y-component)         (a)cannonball's velocity when it leaves the catapult is = U = 48 m/s at an angle of = = 40° (with respect to the horizontal) cannonball is = h = 7.0 m ( above the ground at this time) (a) maximum Height above the ground reached by the cannon ball is Hmax = h + U2 sin2 / 2 g           = 7.0 + (48)2 sin2 (40) / 2 (9.8)           = 55.56 m (b) cannonball makes it over the castle walls and lands back down on the ground, at a horizontal distance from its release point will it land is = R = U2 sin2/ g                                = (48)2 sin2 (40) / 9.8                                = 231.53 m (c) x- and y-components of the cannonball's velocity just before it lands ( The y-axis points up)                            U cos40= 36.77m/s (x-component) U sin40= 30.85m/s   (y-component)         U cos40= 36.77m/s (x-component) U sin40= 30.85m/s   (y-component)

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