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A ball is thrown upward from the roof of a
building 100 m tall with an initial velocity
of 20 m/s. When will the ball reach a height
of 80 m? Please show work.
The formula is
h = hO + vOt - 4.9t²
where
h = final height in m, which here is 80m
hO = initial height in m, which here is 100m
vO = initial velocity in m/s, which here is +20m/s
(+ because it's upward)
t = final time in s, which is what we need to find.
h = hO + vOt - 4.9t²
80 = 100 + 20t - 4.9t²
4.9t² - 20t + 80 = 100
4.9t² - 20t - 20 = 0
49t² - 200t - 200 = 0
Use the quadratic formula:
______
-b ± Öb²-4ac
x = —————————————
2a
where a = 49; b = -200; c = -200
___________________
-(-200) ± Ö(-200)²-4(49)(-200)
x = ————————————————————————————————
2(49)
___________
200 ± Ö40000+39200
x = ————————————————————
98
_____
200 ± Ö79200
x = ——————————————
98
_______
200 ± Ö3600·22
x = ———-—————————————
98
__
200 ± 60Ö22
x = —————————————
98
__
20(10 ± 3Ö22)
x = ———————————————
98
10 __
20(10 ± 3Ö22)
x = ———————————————
98
49
__
10(10 ± 3Ö22)
x = ———————————————
49
Using the +
__
10(10 + 3Ö22)
x = ———————————————
49
x = 4.912399445
Using the -
__
10(10 - 3Ö22)
x = ———————————————
49
x = -0.8308667917
which we discard since we can't have negative time.
So the answer rounded to the nearest tenth is
4.9 seconds.
Edwin
