Question

A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. The maximum height is 0.816 m. The time to achieve the maximum height is 0.985 seconds. B. From the height h in the previous section A, the ball then drops back to y = 0. Find the time t0 that it takes, and the velocity there (magnitude and direction). Show the final formulas for the quantities, along with the numerical answers. C. Find the two times the ball is at the height y = 1/2 h. Show the final formula and the numerical answers.

          A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. The maximum height is 0.816 m. The time to achieve the maximum height is 0.985 seconds.
B. From the height h in the previous section A, the ball then drops back to y = 0. Find the time t0 that it takes, and the velocity there (magnitude and direction). Show the final formulas for the quantities, along with the numerical answers.
C. Find the two times the ball is at the height y = 1/2 h. Show the final formula and the numerical answers.
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. The maximum height is 0.816 m. The time to achieve the maximum height is 0.985 seconds. B. From the height h in the previous section A, the ball then drops back to y = 0. Find the time t0 that it takes, and the velocity there (magnitude and direction). Show the final formulas for the quantities, along with the numerical answers. C. Find the two times the ball is at the height y = 1/2 h. Show the final formula and the numerical answers.
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Transcript

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00:01 Hello, here a ball is thrown vertically with an initial speed of 4 meters per second.
00:07 And first we have to find maximum height.
00:10 Let's do this.
00:10 So let's make this cache.
00:22 This motion occurs only under gravity, that's why maximum height is v0 squared over 2g, which is 0 .816 meters.
00:43 Now we have to determine the time which is needed to reach the maximum height.
00:51 That is v0 over g.
01:05 That's 0 .408 seconds.
01:08 Now in question b we have to actually here we need to label this height as h instead of maximum height.
01:28 So now in question b the object is dropped from this height and first we have to find t0 which is needed to reach the ground so here y coordinate is a function of time is h minus gt squared over 2 therefore this time t 0 is square root of 2 h over g which is 0 .40 8 seconds and now we have to find the final velocity that is negative g t which is negative for meters per second...
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