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A cannonball is fired with an initial velocity of 20 $\mathrm{m} / \mathrm{s}$ and a launch angle of $45^{\circ}$ at a wall 30 $\mathrm{m}$ away. If the cannonball just barely clears the wall, what is the maximum height of the wall?(A) 5.92 $\mathrm{m}$(B) 6.34 $\mathrm{m}$(C) 7.51 $\mathrm{m}$(D) 8.32 $\mathrm{m}$

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Chapter 4

Practice Test 4

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the maximum height of the wall for a project I'll reach, uh can be found by, um using relation. New telly. Creation of motion. That is a delta. Why? Maximum height off the wall can be written in a zoo. Initial we lost e the y component times t minus one word too. Ah g T square, where time is unknown. So we'll find time using Russian horizontal distance. D X is equal to what isn't totally unusual. Lost E times t we console for tea here. The X we have here is a 30. Where is Bea? Nor X We have he's 20 times course off. 45 course for 45 excuse us a time, which is a two 0.12 So now we have time will plunge this time. Really? Here and here. Uh, find the maximum height off the wall, which will become by substituting The values we have is 20. Oh, Sign signed 45 for there. Vertical component of the lost E. Times TV Just 2.12 minus one too. Jeez, a 10. So and then we have time. 2.12 again square. So simplifying this we get the maximum height, which is 7.51 So our Ansari's see end off the problem. Thank you for watching.

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