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uranmaximum [27]
3 years ago
11

PLEASE HELP ILL MARK BRAINLIEST!!! A ball is initially thrown downwards with an initial speed of 20 m/s from the top of a 300 m

tall building. The ball has a constant downward acceleration of 10 m/s ^ 2 a). What is the velocity of the ball immediately before it hits the ground at the base of the building?
Physics
1 answer:
Sholpan [36]3 years ago
8 0

Using the 3rd equation of motion:

= v² - u² = 2gs ------ [g = Acceleration due to gravity]

= v² - 20² = 2 × 10 × 300

= v² - 400 = 6000

= v² = 6000 - 400

= v = √5600

= v = 74.83 m/s

And yeah it's done :)

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Answer:

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Explanation:

The electrostatic force between two charged objects is given by Coulomb's Law:

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Now, when the charges and distance altered as follows:

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using equation (1):

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