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Ball A is thrown straight up with an initial speed v0 and reaches a maximum elevation h before falling back down. When A reaches its maximum elevation, a second ball is thrown straight upward with the same initial speed v0. At what height, y, will the balls cross paths?A. y = hB. y > h/2C. y = h/2D. y < h/2E. y =0

Respuesta :

hamzaahmeds

Answer:

B. y > h/2

Explanation:

The first ball starts at the top with no initial speed. The

second ball starts at the bottom with a large initial speed.

Since, the balls travel the same time until they meet, the

second ball will cover more distance in that time, which

will carry it over the halfway point before the first ball can

reach it.

MrRoyal

The height where the balls cross path is y > h/2

The given parameters are:

Initial speed = v0

When a ball is thrown upward, it follows a parabolic path.

Such that it reaches a maximum height h and then descend downward

From the question, we understand that the ball B is thrown upward when ball A reaches a maximum height

This means that, the two balls would meet when ball A starts descending at a height less than the maximum height.

So, the second ball would cover more distance in that time because it starts at the bottom with the same initial speed (but the speed of B is  greater than the current speed of A)

Hence, the height where the balls cross path is y > h/2

Read more about maximum heights at:

https://brainly.com/question/16653126

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