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 49 m/s at an angle of 45° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time.

(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

v0 = 49 m/s ; =450 ; y0 = 7 m ( above the ground ) ; g = 9.8 m/s2

vector componen of velocity as :

v0x = v0 cos = 49 cos 450 = 34.65 m/s

v0y = v0 sin = 49 sin 450 = 34.65 m/s

movement cannon ball on Y axis :

the cannon ball land on ground it mean that y = 0 m and we can use equation

y = y0 + v0y.t - 0.5.g.t2

0 = 7 + 34.65.t -0.5(9.8).t2 solve this quadratic equation we get t = 7.27 s

movement cannon ball on X axis :

B) x = v0x.t = 34.65. 7.27 = 251.91 m

C) vx = v0x = 34.65 m/s

     vy = v0y - gt = 34.65 - 9.8(7.27) = -36.596 m/s ( down ward )

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