If a rock is dropped from a height of 66 ft, its poistion t seconds after it is dropped until it hits the ground is given by the function s(t)=−16t^2+66 .
Round values below to 3 decimal places.
How long does it take the rock to hit the ground?
Find the average velocity of the rock from when it is released until when it hits the ground.
______feet per second
What time after the rock is thrown will its instantaneous velocity be equal to its average velcity? (Apply the Mean Value Theorem)
___seconds after it is thrown
really? They told you that its height after t seconds is
h = −16t^2+66
so, how long does it take to hit the ground? (height = 0)
16t^2 = 66
t^2 = 66/16
t = 1/4 √66 ≈ 2.031 s
you also know that speed = distance/time, so its average velocity is
-66ft/(1/4 √66)s = -264/√66 ≈ -32.496 ft/s
you know that since h = −16t^2+66, v = -32t
so v = avg velocity when
-32t = -264/√66
t = 264/(32√66) ≈ 1.015 seconds
Note that 1.015 < 2.031 as guaranteed by the MVT.
solve −16t^2+66 = 0
use that to get avg velocity: -66/t
find where -32t = avg velocity
If I understood what this meant, I clearly would have the answer already.
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