Let us suppose that one of the fireworks is launched from the top of the building with an initial upward velocity of 125 ft/sec (correct velocity) and the building has a height of 40 feet. When, in seconds s, will the firework land if it does not explode? Give answer to a tenth of a decimal (ex: 12.7 sec)
h(t) = -16t^2 + 125t + 40
solve using quadratic formula
To find the time when the firework lands, we need to determine when the height of the firework is equal to zero. We can set the equation h(t) = 0 and solve for t using the quadratic formula.
The equation is given by:
h(t) = -16t^2 + 125t + 40
Setting h(t) = 0:
-16t^2 + 125t + 40 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation:
a = -16
b = 125
c = 40
Plugging these values into the quadratic formula, we get:
t = (-125 ± √(125^2 - 4*(-16)*40)) / (2*(-16))
Simplifying:
t = (-125 ± √(15625 + 2560)) / (-32)
t = (-125 ± √(18185)) / (-32)
Calculating the square root and simplifying further:
t = (-125 ± 134.91) / (-32)
Using both the positive and negative values, we have two possible times:
t1 = (-125 + 134.91) / (-32) ≈ 0.294 seconds
t2 = (-125 - 134.91) / (-32) ≈ -7.169 seconds
Since time cannot be negative in this context, the firework will land approximately 0.294 seconds after being launched.