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An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 m away at an average speed of 3.00 m/s, returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?

Answer :

MechEngineer

Answer:

a) 17.97 m/s

b) 16.5 m

Explanation:

The total distance that she runs is twice the distance between her and the table, which is 5.5 * 2 = 11m

So the time she would need to run that distance at a rate of 3m/s is

t = 11/3 = 3.67s

This is also the time it takes for the ball to be thrown up and down, which is twice the time it needs to reach the top, its maximum height at v = 0 m/s

So the time for the ball to reach maximum height is 3.67 / 2 = 1.83 s

Let g = 9.8m/s2. For the ball to get to 0 within 1.83s at the deceleration of 9.8m/s. Its initial speed must be

v = gt = 9.8*1.83 = 17.97 m/s

As she reaches the table within half of the total time (3.67 /2 = 1.83s), the position of the ball would be

[tex]h = vt + gt^2/2 = 17.97*1.83 - 9.8*1.83^2/2 = 16.5 m[/tex]

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