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.
