A blue ball is thrown upward with an initial speed of 19.6 m/s, from a height of
ID: 1405035 • Letter: A
Question
A blue ball is thrown upward with an initial speed of 19.6 m/s, from a height of 0.7 meters above the ground. 2.4 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10 m/s from a height of 21.7 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.
What is the maximum height the blue ball reaches?
How long after the blue ball is thrown are the two balls in the air at the same height?
Explanation / Answer
total time for upward flight of blue ball = 19.6/98 = 2sec so height covered = 19.6m +0.7 =20.3m now height after 0.4 sec of downward journey = 0.78, total height remaining =19.5m v =9.8 x 0.4 =3.6m/s thus they will meet at time t when 10t+ 4.9t^2 = 3.6t + 4.9t^2 + (21.7 - 19.5) t= 11/32 so total time = 2.4 +11/32 =2.75 sec
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