SOLUTION: Using a powerful air gun, a steel ball is shot vertically upward with a velocity of 80 meters per second, followed by another shot after 5 sec. Find the initial velocity of the sec

Question 950010: Using a powerful air gun, a steel ball is shot vertically upward with a velocity of 80 meters per second, followed by another shot after 5 sec. Find the initial velocity of the second ball in order to meet the first ball 150 meters from the ground.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Using a powerful air gun, a steel ball is shot vertically upward with a velocity of 80 meters per second, followed by another shot after 5 sec. Find the initial velocity of the second ball in order to meet the first ball 150 meters from the ground.
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Using 5 m/sec/sec as gravity influence:
For the 1st ball:
h(t) = -5t^2 + 80t
For h(t) = 150
-5t^2 + 80t = 150
t^2 - 16t + 30 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=136 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 13.8309518948453, 2.1690481051547. Here's your graph:

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t = 2.169 seconds ascending, less than 5 seconds --> not possible for them to meet.
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t = 13.831 seconds descending at 150 meters.
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The 2nd ball has to get to 150 meters in 8.831 seconds (13.831 - 5)
h(t) = -5t^2 + v*t
150 = -5*8.831^2 + v*8.831
8.831v = 539.9328
v =~ 61.1406 m/sec

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