28.7 = 28.7 t=> t = 1 sec =================================Verification:The ball thrown downwards covers distance in 1 sec= (1/2)gt^2 = (1/2) * 9.81 * (1)^2 = 4.905 mThe ball thrown upwards with velocity 28.7 m/s covers distance in 1 sec= 28.......
If H = distance covered downwards by the ball dropped when the two balls meet,
then 28.7 - H = distance covered by the ball thrown upwards
H = (1/2)gt^2 and
28.7 - H = (23.7)t - (1/2)gt^2
Adding,
28.7 = 28.7 t
=> t = 1 sec <== Answer
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Verification:
The ball thrown downwards covers distance in 1 sec
= (1/2)gt^2 = (1/2) * 9.81 * (1)^2 = 4.905 m
The ball thrown upwards with velocity 28.7 m/s covers distance in 1 sec
= 28.7 * 1 - (1/2)*(9.81)*(1)^2
= 28.7 - 4.905 m
Sum of these distances = 28.7 m.
One can take any direction as positive, but then for all vector quantities like velocity, displacement and acceleration due to gravity, the same convention should be followed. You took g as negative means you considered upward direction as positive. In that case, H should be taken as negative.
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Find y(t) = h - Uyt - 1/2 g^2 = Vy t - 1/2 gt^2 = Y(t) meaning both balls are at the same height at time t; where h = 28.7 m, Uy = 0, initial drop speed from the building, Vf = Vy = sqrt(2gh) from the ground. Find t.
h = Vy t = sqrt(2gh) t so t = h/sqrt(2gh) = 28.7/sqrt(2*9.8*28.7) = 1.21 seconds. ANS.
I don't see anywhere in your work where you properly found Vf based on that first drop, which becomes Vy the upward toss speed.
