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eznaoum
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$15.00 Physics HELLPPPPPPPPPPPPPPP

From Physics: General-Physics
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Due on Jan. 31, 2009
Asked on Jan 31, 2009 at 4:05:08PM

Q:
A major leaguer hits a baseball so that it leaves the bat at a speed of 28.5 m/s and at an angle of 35.7 degrees above the horizontal. Ignore air resistance.
1.At what two time is the baseball 9.5m above the point at which it left the bat?
2.Calculate the horizontal component of the baseball's velocity at each of the two times.
3.Calculate the vertical component at each of the two times.
4.What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?
5.What is the direction of the baseball's velocity when it returns to bat level?

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gorgo20
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Posted on Jan 31, 2009 at 4:54:43PM

A:
Preview: ... n is constant, equal to 23.144mps 3. at time 2.667 the vertical ve ...

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sudipta

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Posted on Jan 31, 2009 at 5:18:17PM

A:
Preview: ... wo times = 28.5 cos(35.7 degrees) m/s = 28.5 * 0.8121 m/s = 23.1444 m/s 3. The vertical component of the baseball's velocity at 0.727 seconds = [16.6309 - 9.8 * 0.727] m/s = 9.507 m/s The vertical component of the baseball's velocity at 2.667 seconds = [16.6309 - 9.8 * 2.667] m/s = -9.507 m/s 4. When the baseball returns to the level at which it left the bat, the displacement S = 0 m Suppose the time taken i ...

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mellisa

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Posted on Jan 31, 2009 at 6:32:33PM

A:
Preview: ... ity at each of the two times. 3.Calculate the vertical component at each of the two times. 4.What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat? 5.What is the direction of the baseball's velocity when it returns to bat leve ...

The full tutorial is about 195 words long .