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vlabodo
7 days ago
11

Suppose Sammy Sosa hits a home run which travels 361. ft (110. m). Leaving the bat at 50 degrees above the horizontal, how high

did it travel?

Physics
1 answer:
Yuliya22 [2.4K]7 days ago
5 0

Response:

The horizontal span of Sosa is 276.526 ft or 84.28 meters.

Explanation:

As illustrated in the diagram, let point O denote Sosa's starting position. She travels 361 ft at a 50-degree angle relative to the horizontal.

sin 50 = \frac{OM}{OP}

0.7660 = h / 361

h = 276.526 ft

h = 84.28 meters

The horizontal distance of Sosa is 276.526 ft or 84.28 meters.

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Answer:

Speeds of 1.83 m/s and 6.83 m/s

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Based on the law of conservation of momentum,

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2*(5^{2})= v_1^{2} + v_2^{2}We know that v_1+v_2=5 leads to v_1=5-v_2

50=(5-v_2)^{2}+ v_2^{2}

50=25+v_2^{2}-10v_2+v_2^{2}

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The answer is:

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