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 35 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?
_____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?
____m

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

Explanation / Answer

Given data : vo = 35 m/s   ; = 40o   ; yo = 7 m

A) ymax = ( vo sin )2/2g = ( 35 sin 40 )2/(2.9.8) = 25.82 m

maximum height reached = Y = yo + ymax = 7 + 25.82 = 32.82 m

B) y = yo + vo sin .t - 0.5g.t2         when reach ground y = 0

0 = 7 + 35 sin 40o.t - 0.5.(9.8).t2

0 = 7+ 22.498t - 4.9t2

solve this equation get : t = 4.884 s

X = vo cos .t = 35.cos 40o.(4.884) = 130.948 m

C) vx = vo cos 40 = 26.811 m/s

     vy = vo sin 40o - g.t = 35 sin 40 - 9.8.(4.884) = -25.366 m/s

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