SOLUTION: THIS IS A TWO PART QUESTION h(t)=-16t^2+vt+h you launch a firework from the ground with velocity 70ft per second calculate the maximum height if it is set to explode 3 seco

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: THIS IS A TWO PART QUESTION h(t)=-16t^2+vt+h you launch a firework from the ground with velocity 70ft per second calculate the maximum height if it is set to explode 3 seco      Log On


   

Question 1009692: THIS IS A TWO PART QUESTION
h(t)=-16t^2+vt+h
you launch a firework from the ground with velocity 70ft per second calculate the maximum height if it is set to explode 3 seconds after launch.
you launch another firework from the ground set to explode at 142ft with initial velocity 56ft per second how long after setting off the firework should the delay be set
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
h(t)=-16t^2+vt+h
you launch a firework from the ground with velocity 70ft per second calculate the maximum height if it is set to explode 3 seconds after launch.
===================
Its max height is not affected by the 3 second timer.
h(t) = -16t^2 + 70t
The max ht is the vertex, on the Line of Symmetry.
The LOS is t = -b/2a = -70/-32
Max ht = -16(35/16)^2 + 70*35/16
= 76.5625 feet
---------------------
you launch another firework from the ground set to explode at 142 ft with initial velocity 56 ft per second how long after setting off the firework should the delay be set
h(t) = -16t^2 + 56t = 142
-16t^2 + 56t - 142 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B56x%2B-142+=+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=%2856%29%5E2-4%2A-16%2A-142=-5952.

The discriminant -5952 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -5952 is + or - sqrt%28+5952%29+=+77.1492060879436.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B56%2Ax%2B-142+%29

=============
t = no real number solution.
----
The max ht is 49 feet, it never gets to 142 feet.