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A ball is thrown in a projectile motion trajectory with an initial velocity $v$ at an angle $\theta$ above the ground. If the acceleration due to gravity is $-g,$ which of the following is the correct expression of the time it takes for the ball to reach its highest point, $y,$ from the ground?(A) $v^{2} \sin t / \mathrm{g}$(B) $-v \cos \theta / g$(C) $v \sin \theta / g$(D) $v^{2} \cos \theta / g$

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Chapter 12

Practice Test 2

Section 1

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this problem, we have given a ball is thrown in a projectile motion trajectory with an initial speed. We at an angle theater of the ground. This has gone the same well is trita have the ground. If the excavation due to gravity is minus G. We know that in the downward direction. Which of the following is the correct expiration of the time it takes for the ball to reach its highest point. Right? So we know that this is supposed highest point. Right? So we have to find this time. So we know that total time. We can find from this point from point A. To point B. We can find the time and this time will be equal to the half of that time was light. So the total time was light. We can find it is a call to We know that as y is equal to you White B. Let's have where it is where we will use this kind of attack occasion. So first we will find that total time of light says when violence land on the ground, The vertical displacement. Will we call to zero from point a to week vertical displacement 20. And initially speedy's recent data into feet and this is half into G. P square. So we can find the their time. It is a cool do he is a call to to be signed by G. This is the total time of slide. Great. No. We have to find the time from the initial point to the highest point. So this will be equal to the half of this time. So our time strong initial point to the maximum point where we called T baidu or we can say the scientific obliging. Now we will check the options. Option C. Is correct. We can write option C. Is correct.

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