A cannonball is catapulted toward a castle. The cannon-ball\'s velocity when it
ID: 1957010 • Letter: A
Question
A cannonball is catapulted toward a castle. The cannon-ball's velocity when it leaves the catapult is 40m/s at an angle of 37 degree with respect to the horinzontal and the cannonball is 7.0m 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 horinzontal distance from its release point will it land?
c). What are the x- and y- components of the cannonbal's velocity just before it lands? The y-axis points up.
Explanation / Answer
some of the velocity is straight up and some is towards the castle when it leaves the tip of the barrel. 37 degrees is small (less than 45), so more of the velocity will be toward the caslte than up. The trig function (choice between cos and sin) that is smaller for small angle is sin 40m/s sin(37) = 24.1m/s is the vertical component of the velocity. (note that more than half of the speed is vertical and that more than half is horizontal. This is how sides of right triangles work) Now we can use one of our kinematic equations and find displacment when bal stops going up. Vf^2 = Vo^2 + 2ad Vf = 0 when it stops going up Vo^2 = 2gd (since a is in oppositie direction from Vo vert) Vo^2/2g = d = 579.5m^2/s^2/(2*9.81m/s^2) = 29.5 m above where it left cannon. It left cannon 7m up so: max height = 29.5 m + 7m = 36.5 m ANSWER to part a b) up and down and side to side are not related except they both happen for the same amount to time. Let's find time for cannon ball to go up and come down. For a ball to go up 29.5 m takes its average speed times time. We can use distance = rate times time d = rt d/r = t average rate is sum of speed at bottom plus speed at top divided by two 29.5m/{(24.1m/s + 0m/s)/2} = 2.45s it takes 2.45s to go up on the way down it goes 7m further. d = vot + (1/2)at^2 change in height is down to the ground, gravity is pulling masses down, so both are in the same direction. t^2 = 2d/a t = SQRT(2d/a) t = SQRT(2*36.5m/9.81m/s^2) = 2.73 s note that it takes a little longer to fall the extra 7m Time in the air is time to go up plus time to come down = 5.18s this whole time the horizontal speed is constant Horizontal distance = rt = 40m/s cos(37) *5.18s = 165.6m c) it is going a little faster downward when it lands due to falling the extra 7m, but the x component is what it has been the whole time (neglecting wind resistance) x component = 40m/s cos(37) = 31.9m/s y component can be found using d = rt average speed on way down = r = d/t = 36.5m/2.73s = 13.4m/s average speed = (Vf + Vo)/2, but Vo at peak of trajectory is 0m/s so Vf = V ave *2 = 26.8 m/s y component = - 26.8m/s ANSWER x 31.9m/s y -26.8m/s
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