Question

A red ball is thrown down with an initial speed of 1.1m/s from a height of 28 meters above the ground. Then 0.5 seconds after the red ball is thrown, a blue bar) is thrown upward with an initial speed of 24.6m/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 9.81m/s^2
1. What is the speed of the red ball right before it hits the ground?
2. How long does it take the red ball to reach the ground?
3. What is the maximum height the blue ball reaches?
4. What is the height of the blue ball 2 seconds after the red bal is thrown
5. How long after the red ball is thrown are the two balls in the air at the same height?

Answer 1

1.
v2 = u2 + 2gh

v2 = 1.12 + 2 * 9.8 * 28 = 550.01

v = 23.45 m/s

The speed of the red ball right before it hits the ground = 23.45 m/s.

2.

v = u + gt

23.45 = 1.1 + 9.8 * t

t = 2.28 seconds

The time it take the red ball to reach the ground = 2.28 seconds,

3.

Height above the ground = 0.8 m

v2 = u2 - 2gh

0 = 25.32 - 2* 9.8* h

h = 32.66 m.

The maximum height the blue ball reaches = (h + 1) = (32.66 + 1) = 33.66 m.

4.

Time of travel of the blue ball = (1.9 - 0.5) = 1.4 seconds.

s = ut - 0.5 gt2

s = 25.3* 1.4 - 0.5 * 9.8 * 1.42= 25.816 m.

The height of the blue ball 1.9 seconds after the red ball is thrown = (s + 1) = (25.816 + 1) = 26.816 m.

5.

Let the time of travel of the red ball be t seconds.

So the time of travel of the blue ball = (t - 0.5) seconds.

Both the balls are at the same height :

28 - (s) = 1 + (h) ................{"s" & "h" are the displacements of the red & the blue ball respectively.}

28 - (ut + 0.5 gt2) = 1 + (ut - 0.5gt2)

28 - (1.1 t + 0.5 * 9.8 t2) = 1+ (25.3 (t-0.5) - 0.5*9.8*(t-0.5)2)

Now we have to solve the above equation to find the time after which both the balls are at the same height.

28 - 1.1t - 4.9t2 = 1 + 25.3t - 12.65 - 4.9t2 + 4.9 t - 1.225

(28 - 0.8 + 12.65 +1.225) = (25.3 + 4.9 + 1.1) * t

t = 41.075 / 31.3 = 1.3123 = 1.31 seconds (approx.)

The time after the red ball is thrown are the two balls in the air at the same height = 1.31 seconds.