solve −16t^2+66 = 0
use that to get avg velocity: -66/t
find where -32t = avg velocity
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?
_____seconds
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
3 answers
Explain more?
If I understood what this meant, I clearly would have the answer already.
If I understood what this meant, I clearly would have the answer already.
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.
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.
3 answers