Question 1009692
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.
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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
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 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
*[invoke solve_quadratic_equation -16,56,-142]
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t = no real number solution.
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The max ht is 49 feet, it never gets to 142 feet.