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Alexxandr [17]
3 years ago
6

g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th

e ball's initial speed (in m/s).
Physics
1 answer:
kati45 [8]3 years ago
3 0

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

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Answer:

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Explanation:

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