The height (in meters) of a shot cannonball follows a trajectory given by h(t) = -4.9t^2 + 18t - 0.4 at time t (in seconds). As an improper fraction, for how long is the cannonball above a height of 4 meters?
+128473 Let's look at the times when it is at 4m
-4.9t^2 + 18t - .4 = 4
-4.9t^2 + 18t - .4 - 4 = 0
-4.9t^2 + 18t - 4.4 = 0 multiply through by -10
49t^2 - 180t + 44 = 0
Using the Quadratic Formula
180 ± sqrt [ 180^2 - 4(49) (44) ] 180 ± sqrt [ 23776]
t = ____________________________ = _________________
2 (49) 98
The time when it is above 4m is given by
sqrt [ 23776] / 98 - - sqrt [ 23776] / 98 =
2sqrt [ 23776 ] / 98 =
sqrt [ 23776] / 49 sec
+307 Answer: \(4\sqrt{1486}\over{-49}\)
Solution:
h(t) should be equal to 4.
So:
-4.9t² + 18t - 0.4 = 4
-4.9t² + 18t - 4.4 = 0
Putting this into the quadratic formula gives:
\(-18\pm\sqrt{237.76}\over{-9.8}\)
You need the bigger one of these roots minus the smaller one, which is equal to 2 times the square root of 237.76 over -9.8. This is also equal to \(4\sqrt{1486}\over{-49}\).
