A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0

Question

A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.5 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.4 m/s from a height of 28.9 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?

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

A

B

C

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

The acceleration is positive and it is speeding up

The acceleration is negative and it is speeding up

The acceleration is positive and it is slowing down

The acceleration is negative and it is slowing down

Explanation / Answer

1)
At maximum height speed is alwayd 0 m/s
Answer: 0m/s

2)
Lets say they meet at height h above the ground.
now calculate time at which each ball to reach height h
Lets take upwards direction as positive an ddownward direction a snegative.

using:
d = do + vi*t - 0.5*g*t^2

For blue ball:
h = 0.5 + 23t - 0.5*9.8*t^2

For red ball:
h = 28.9 + ( - 6.4)(t-2.8) - 0.5*9.8*(t-2.8)^2

equating both h we get,
0.5 + 23t - 0.5*9.8*t^2 = 28.9 + ( - 6.4)(t-2.8) - 0.5*9.8*(t-2.8)^2
0.5 + 23t - 4.9*t^2 = 28.9 - 6.4 t + 17.92 - 4.9* (t^2-5.6t +7.84)
0.5 + 23t - 4.9*t^2 = 28.9 - 6.4 t + 17.92 - 4.9t^2 + 27.44t - 38.416
0.5 + 23t = 28.9 - 6.4 t + 17.92 + 27.44t - 38.416
1.96 t = 7.904
t = 4 s
Answer : 4s

3)
Both the curves must be parabolic.
Answer: C

4.
Answer: The acceleration is positive and it is speeding up

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