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Cannonball Launch!

  • #1

Homework Statement



A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 32 m/s at an angle of 50° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time.

a) 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?


b)What is the y-component of the cannonball's velocity just before it lands? The y-axis points up.

For this question, I did Vy=32sin50=24.5 (but this was also incorrect)


Homework Equations



R= lvl^2/9 * (sin2theta)



The Attempt at a Solution



For part a, I used the equation for Range and got 102.9m but it was incorrect.

For part b, I did Vy=32sin50=24.5 (but this was also incorrect)
 

Answers and Replies

  • #2
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The determination of the initial vertical velocity of 24.5 m/s is correct. Based on this, the vertical position at any time t is given by $$s=7.0 +24.5t-4.9t^2$$ The time for the cannonball to hit the ground would be the time at which s = 0. Solving the quadratic equation yields t = 5.27 sec. The horizontal velocity at time zero is 32 cos 50 = 20.57 m/s. So the range is 20.57 x 5.27 = 108 m.
 

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