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 61 m/s at an angle of 37° with respect to the horizontal and the cannonball is 5.0 m 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 horizontal distance from its release point will it land? (c) What are the x and y-components of the cannonball's velocity just before it lands? The y-axis points up.

a) The maximum height reached is 70 m.

b) The cannonball lands 165 m from its release point.

c) The x-component velocity is 49 m/s, and the y-component is m/s.

Explanation / Answer

Given that

The cannonball's velocity when it leaves the catapult is(v) = 61 m/s

The cannonball leaves at an angle with the horizontal is (theta) = 37°

The cannon ball at a height if = 5.0 m above the ground at this time.

a)

The maximum height reached above the ground is given by

Hmax =h+v2sin2theta/2g=5+(61)2sin2(37)/2*9.81=5+68.688 =73.688m

b)

The horizontal distance from its release point will it land is R =v2sin2theta/g =(61)2sin2*(37)/9.81 =364.613m

c)

The x-component of velocity is vx =vcostheta =61*cos(37) =48.716m/s

The y-component of velocity is vy =vsintheta =61*sin(37) =36.7107m/s

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