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 39 m/s at an angle of 30° 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

Vo=39 m/s==>vox=Vo*cos(30o)39*3/2 m/s==>voy=Vo*sin(30o)39/2 m/s

max ht=y0+voy2/(2*g)7+(39/2)2/(2*9.8)26.4 m max ht==>max ht t=voy/g

a) 26.4 m

-y=voy*t-1/2*9.8*t2=5.0471 s==>x=vox*t==>39*3/2*5.0471=170.5 m=range

b) x=vox*t=170.5 m

vyfinal=voy-g*t=39/2-9.8*5.0471=-30 m/s

c) vxfinal=vox=39*3/2 m/s==>vyfinal=voy-g*t=39/2-9.8*5.0471= -30 m/s

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