h is the height of the ball and t is time. Since we want to find the time when the ball touches the floor, then height is 0. This leaves us with the equation

-16 $t^{2}$ + 4t + 4 = 0

This is a quadratic equation can be solved with the following formula:

$x= \frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

where a=-16

b=4

c=4

Solving for t we will find two different results:

$t1=\frac{-4-\sqrt{272} }{2(-16)} =0.125+0.125\sqrt{17} =0.64039$ $t2=\frac{-4+\sqrt{272} }{2(-16)} =0.125-0.125\sqrt{17} =-0.39039$

Since time can't be negative, we discard t2 and choose t1.

Since it is required to answer in the nearest hundredth, we round the result to t=0.64 seconds.

5 0

2 years ago

During batting practice, two pop flies are hit from the same location, 2 s apart. The paths are modeled by the equations h = -16

jolli1 [7]

To find the time at which both balls are at the same height, set the equations equal to each other then solve for t.
h = -16t^2 + 56t
h = -16t^2 + 156t - 248
-16t^2 + 56t = -16t^2 + 156t - 248
You can cancel out the -16t^2's to get
56t = 156t - 248
=> 0 = 100t - 248
=> 248 = 100t
=> 2.48 = t
Using this time value, plug into either equation to find the height.
h = 16(2.48)^2 + 56(2.48)
Final answer:
h = 40.4736
Hope I helped :)

6 0

2 years ago

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