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) About Me  (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 ax%5E2%2Bbx%2Bc=0 (in our case -4.9x%5E2%2B2.3x%2B3.5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%282.3%29%5E2-4%2A-4.9%2A3.5=73.89.

Discriminant d=73.89 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2.3%2B-sqrt%28+73.89+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282.3%29%2Bsqrt%28+73.89+%29%29%2F2%5C-4.9+=+-0.6424417621568
x%5B2%5D+=+%28-%282.3%29-sqrt%28+73.89+%29%29%2F2%5C-4.9+=+1.11182951725884

Quadratic expression -4.9x%5E2%2B2.3x%2B3.5 can be factored:
-4.9x%5E2%2B2.3x%2B3.5+=+%28x--0.6424417621568%29%2A%28x-1.11182951725884%29
Again, the answer is: -0.6424417621568, 1.11182951725884. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-4.9%2Ax%5E2%2B2.3%2Ax%2B3.5+%29

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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)