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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?

Aug 3, 2022

#1
+129803
+2

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

Aug 3, 2022
#2
+308
+3

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 $$4\sqrt{1486}\over{-49}$$.

Aug 3, 2022