Question

A blue ball is thrown upward with an initial speed of 20.8 m/s, from a height of 0.9 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 8 m/s from a height of 25.4 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.

Answer 1

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

Answer 2

I had an interesting idea on how to solve this but I don't have time to post it. Does anyone have any ideas?

Answer 3

\[downward acceleration\neq 9.81 but = 9.86\]

Answer 4

the question itself is wrong

Answer 5

Time taken for B ball to reach top: 20.8/9.81 = 2.12 sec
The top = 0.9 + 20.8*2.12 - 0.5*9.81*2.12^2 = 22.9 meter
When R ball is thrown, B ball has traveled 1/2*g(2.5-2.12)^2 from its top, or 0.5*9.81 *(2.5-2.12)^2 = 0.71 meters (from its top), or (22.9-0.71) = 22.24 meters from the ground.

The moment B ball reaches 22.24 meters from the ground, R ball is thrown. When B ball and R ball have the same height (h-B = h-R):
height of B ball = 22.24 - 0.5*g*t^2
height of R ball = 8*t + 0.5*g*t^2
solve this equation with WolframAlpha ( http://www.wolframalpha.com/input/?i=+22.24+-+0.5*9.81*t^2+%3D+8*t+%2B+0.5*9.81*t^2), we get:

t = 1.15 secs

Answer 6

Sorry, actually the time above is calculated since red ball was thrown. For the time since blue ball was thrown: 1.15 + 2.5 = 3.65 secs

Answer 7

I figured it out on my own using with the assistance of mathematica! Retwick, again, please share.

I didn't even write the question. I literally copied it from a physics program used nationally. Please explain what is soooo wrong about the question. Thank you, mlutfi.