SOLUTION: A body is thrown vertically upward from the distance h = 1.5 m above the ground at the edge of a pit having a depth of s = 3.5 m The initial velocity of body is 2.3 m/s Determin

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Question 378930: A body is thrown vertically upward from the distance h = 1.5 m above the ground at the edge of a pit having a depth of s = 3.5 m
The initial velocity of body is 2.3 m/s
Determine the velocity at which the body will reach the bottom of the pit.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
You didn't give the info, so I'll assume this is on the Earth and use
h(t) = -4.9t^2 + vt, t in seconds, h in meters
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h(t) = -16t^2 + 2.3t
Find t for h(t) = -3.5 meters
-3.5 = -4.9t^2 + 2.3t
-4.9t^2 + 2.3t + 3.5 = 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=73.89 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: -0.6424417621568, 1.11182951725884. Here's your graph:

----------------
t = 1.1118 seconds (Ignore the negative solution)
velocity = h'(t) = -9.8t + 2.3
h'(1.1118) = -8.59564 m/sec at impact (negative = going down)

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