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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 14

Practice Test 4

Section 1

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this problem we have given And cannon ball is fired with an initial velocity which is 20 m purse. Again, An angle is 45 degrees at the wall. We just talked him into the way. Yeah, Yeah, like this. We have to find out the height of this wall if can involve just to barely clears the wall. So first of all, we will find that time. So we know that in horizontal direction velocity is V X, which is 20 costs 45 and the stand stable in the general election is 30 m. So time will be called to the stand by velocity. That means activity 20 cost 45 So this time is equal to 20 divided by. So the 30 divided by 20 into costs 45. So this is a call to 2.12 seconds. Now we have to find this height edge Suppose height of Wallace Edge. Now we will use automatic occasion as a is equal to you by t plus half. Hey, by P square, we will apply this can, um, Arctic inclusion in the vertical direction. The vertical displacement is it's called to edge initial velocity in the y direction is 20 and two signed 45 time is given. We have fine time. We just 2.12 Okay, plus half acceleration due to gravity is minus 9.8 in 22.12 square. So this is the height of the wall. Now we will solve this. This is equal to 20. And to sign 45 into 2.1 to minus 0.5, multiply with 9.8 multiply was 2.12 square. So this is equal to 7.9 six m. Now, if we check our options, we have given 5.9 to 6.3 to 47.51 So the best auction will be equal to 7.51 ocean. See will be correct Option. Thank you.

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