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The Question: A boy whirls a stone in a horizontal circle of radius1.1m and at h

ID: 1668312 • Letter: T

Question

The Question:
A boy whirls a stone in a horizontal circle of radius1.1m and at heigh 2.2m above ground level. The stringbreaks, and the stone flies horizontally and strikes the groundafter traveling a horizontal distance of 10m. What is the magnitudeof the centripetal acceleration of the stone while in circularmotion?
I know first to get x and y, but I am running intodifficulties finding them. I know x = vo t and y = -(1/2) g t2 I understand how to finish out the problem, but for somereason am having trouble with calculating the x and y values. Any help will be appreciated. Thanks! The Question:
A boy whirls a stone in a horizontal circle of radius1.1m and at heigh 2.2m above ground level. The stringbreaks, and the stone flies horizontally and strikes the groundafter traveling a horizontal distance of 10m. What is the magnitudeof the centripetal acceleration of the stone while in circularmotion?
I know first to get x and y, but I am running intodifficulties finding them. I know x = vo t and y = -(1/2) g t2 I understand how to finish out the problem, but for somereason am having trouble with calculating the x and y values. Any help will be appreciated. Thanks! I understand how to finish out the problem, but for somereason am having trouble with calculating the x and y values. Any help will be appreciated. Thanks!

Explanation / Answer

ac (centripetal) = v2/r You know that the time it took to hit the ground isfound using: y = -(1/2) g t2 2.2 = (.5*9.8)t2 (The negative isunimportant to this problem) Find "t" Time is constant so we use that for the x directionalso. x = v0t +1/2(a)(t)2 10 = v0(t) + (0)(t)2 Acceleration in the x direction doesn't change and plug in thet for the value we found above to find the initial velocity. Theinitial velocity is the same velocity as the point right before thestring broke. ac (centripetal) = v2/r We have v and r. v was found above and r = 1.1. Findac (centripetal) Acceleration in the x direction doesn't change and plug in thet for the value we found above to find the initial velocity. Theinitial velocity is the same velocity as the point right before thestring broke. ac (centripetal) = v2/r We have v and r. v was found above and r = 1.1. Findac (centripetal)
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