A red ball is thrown down with an initial speed of 1.4 m/s from a height of 28 m

Question

A red ball is thrown down with an initial speed of 1.4 m/s from a height of 28 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 25.1 m/s, from a height of 1 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 red ball right before it hits the ground?

m/s

2)How long does it take the red ball to reach the ground?

s

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

m

4)What is the height of the blue ball 2 seconds after the red ball is thrown?

m

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

s

6)Which statement is true regarding the blue ball?

After it is released and before it hits the ground, the blue ball is always moving faster than the red ball at any given time.

After it is released and before it hits the ground, the blue ball is sometimes moving faster than the red ball at any given time.

After it is released and before it hits the ground, the blue ball is never moving faster than the red ball at any given time.

7)

Which statement is true about the red ball?

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) The speed of the red ball right before it hits the ground, v^2= u^2+2gh = (1.4)^2 +2* 9.81* 28

v= 23.48 m/s

2) u = 1.4 s = 28 t = ? a = 9.81

s = ut + 1/2at^2

Therefore 4.905t^2 + 14t - 28 = 0

Using your calculator or Quadratic formula you get t = 2.25078.... and t = -2.536208
therefore t = 2.3 seconds

3) Maximum height the blue ball reaches

v = 0.0 m/s
u = 25.1 m/s
a = -9.81 m/s^2 (Earth gravity, acting against the line of motion makes it negative)
t = ?

(0.0 m/s) - (25.1 m/s) = t * (-9.81 m/s^2)

(-25.1 m/s) / (-9.81 m/s^2) = t

time = 2.56 seconds
Solve for distance
s = 0.5 * (v + u) * t
s = 0.5 * [ (0.0 m/s) + (25.1 m/s) ] * (2.56 s)
s = 0.5 * (25.1 m/s) * (2.56 s)
s = 32.13m

4) s= ut+0.5gt^2 = 25.1 *2+0.5 *9.81*2^4 = 69.82 m

5) The height of the red ball is y = 28-1.4t-(1/2)(g)t^2, while the height of the blue ball is y = 0.9+25.1(t-0.6)-(1/2)(g)t^2.

Equating these yields:
28-1.4t-(1/2)(g)t^2 = 0.9+25.1(t-0.6)-(1/2)(g)t^2 --> 28-1.4t = 0.9+25.1(t-0.6) --> t = 1.59 s

6) After it is released and before it hits the ground, the blue ball is never moving faster than the red ball at any given time

7) The acceleration is positive and it is slowing down

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