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A boy wants to throw a can straight up and then hit it with a second can. He wants the collision to occur 4.0 meters above the throwing point. In addition, he knows that the time he needs between throws is 3.0 seconds. Assuming he throws both cans with the same speed, what must the initial speed be

Respuesta :

tushargupta0691

16.03 m/s is the initial speed of the cans.

Newton’s equations of motion has to be used to solve it;

He wants collision at height (h) = 4.0 m

The acceleration due to gravity  as,

                                                      g =9.8ms−²

The initial speed of the thrown can is calculated by:

h = vt - ¹/₂gt²

4 = 3t - (0.5)(9.8)(3)²

4 = 3t - 44.1

3t = 48.1

t = 48.1/3

t = 16.03 m/s

Thus, the initial speed of the boy's thrown can is 16.03 m/s.

By measuring the rate at which the velocity varies with regard to the passage of time, one can calculate the acceleration of a body. By calculating the rate at which the displacement varies with regard to the passage of time, the velocity may be calculated. The perpendicular distance between the body's initial and final locations is its displacement. These are all vector quantities, meaning they all have both magnitude and direction.

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https://brainly.com/question/24493758

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