SOLUTION: During World War 1, mortars were fired from trenches 3 feet down. the mortars had a velocity of 150ft/s use the formula h=-16^2+vt+s to deterrmine how long it will take for the mor

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Question 1012435: During World War 1, mortars were fired from trenches 3 feet down. the mortars had a velocity of 150ft/s use the formula h=-16^2+vt+s to deterrmine how long it will take for the mortar shell to strike its target?
Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
During World War 1, mortars were fired from trenches 3 feet down. the mortars had a velocity of 150ft/s use the formula h=-16^2+vt+s to determine how long it will take for the mortar shell to strike its target?
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You didn't give an angle. Are they firing straight up? They'll be disappointed with the results.
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h=-16^2+vt+s
It's h=-16t^2 + vt + s --- not 16^2
h = -16t^2 + 150t - 3
Solve for t when h = -3, or when h = 0.
For h = 0:
h = -16t^2 + 150t - 3 = 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=22308 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 0.0200428496878517, 9.35495715031215. Here's your graph:

==========
The small value is ground level ascending.
t =~ 9.355 seconds to impact at ground level.


Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
h=-16t^2+150t-3
Set it equal to 0 and look for largest positive root.
-16t^2+150t-3=0
16t^2-150t+3=0
t=(1/32)(150+/- sqrt (22500-192);sqrt (22308)=149.36
t=299.36/32=9.355 (time it reaches target)
t=0.64/32=0.02 sec (time it reaches the ground surface on the way up)
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