SOLUTION: A firework is launched upward at an initial velocity of 49 m/s, from a height
of 1.5 m above the ground. The height of the firework, in metres, after t seconds, is modelled by the
Question 1187125: A firework is launched upward at an initial velocity of 49 m/s, from a height
of 1.5 m above the ground. The height of the firework, in metres, after t seconds, is modelled by the equation h=4.9t^2 + 49t + 1.5.
a) What is the maximum height of the firework above the ground?
b) Over what time interval is the height of the firework greater than 100 m above
the ground? Round to the nearest hundredth of a second. Found 2 solutions by Boreal, ikleyn:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! I assume this should be h(t)=-4.9t^2+49t + 1.5
this is a negative quadratic so the vertex t=-b/2a is the time at maximum, and that is -49/-9.8 or 5 seconds
the maximum height is 124 m.
solve for -4.9t^2+49t+1.5=100
or-4.9t^2+49t-98.5=0=4.9t^2-49t+98.5
t=(1/9.8)(49+/-sqrt (2401-1930.6)); square root=21.69
t=70.69/9.8=7.24 sec
t=27.31/9.8=2.79 sec.
those are the times
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