A blue ball is thrown upward with an initial speed of 23.7 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 23.7 m/s, from a height of 0.9 meters above the ground. 2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.6 m/s from a height of 31.4 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

1) What is the speed of the blue ball when it reaches its maximum height? m/s

2) How long does it take the blue ball to reach its maximum height?

3) What is the maximum height the blue ball reaches?

4) What is the height of the red ball 3.77 seconds after the blue ball is thrown?

5) How long after the blue ball is thrown are the two balls in the air at the same height?

6) Which statement is true about the blue ball after it has reached its maximum height and is falling back down?

A - The acceleration is positive and it is speeding up

B - The acceleration is negative and it is speeding up

C - The acceleration is positive and it is slowing down

D - The acceleration is negative and it is slowing down

Explanation / Answer

s the ball moves upward, its vertical velocity decreases from 23.7m/s to 0 m/s at the rate of 9.8 m/s each second. The ball stops moving upward when its vertical velocity is 0 m/s. At this time the ball is at its maximum height.

Time up = Change of velocity ÷ acceleration

Change of velocity = vf - vi = 0 - 23.7 = -23.7m/s

Acceleration = -9.8 m/s ^2

Time up = -23.7 ÷ -9.8 2.418 seconds

Distance up = vi * t - ½ * 9.8 * t^2 = 23.7 * 23.7/9.8 - 4.9 * (23.7/9.8)^2 = 28.65meters

Maximum height = 28.65 + 0. = 29.55 meters

Next the blue ball accelerates at from 0 m/s to its final velocity as it falls 29.55 meter

Blue balls' travel time = 3.77 - 2.9 = 0.87s to find the height of a ball thrown down with an initial velocity of 7.6 m/s 0.87s after it was thrown define down as the negative direction; then our initial height is 23.7 m, initial velocity is -10.1 m/s and accel = -9.81m/s/s y(t) = y0 + v0 t + 1/2 at^2 y(t) = 23.7 - 7.6 t - 1/2 *9.8 t^2 when t = 0.87s: y(t) = 23.7 - 7.6x0.78 - 1/2*9.8*(0.87)^2 = 14.06 m above the floor

6) A. THE ACCELERATION IS POSITIVE AND IS SPEEDING UP

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