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A rock is tossed from a platform and follows a parabolic path through the air. The height of the rock in meters is given by h(t) = -5t^2 + 12t + 14, where t is measured in seconds.
a. How high is the rock off the ground when it's thrown?
b. How long is the rock in the air?
c. For what times is the height of the rock greater than 17 m?
d. How long is the rock above a height of 17 m?
A rock is tossed from a platform and follows a parabolic path through the air. The height of the rock in meters is given by h(t) = -5t^2 + 12t + 14, where t is measured in seconds.
a. How high is the rock off the ground when it's thrown?
b. How long is the rock in the air?
c. For what times is the height of the rock greater than 17 m?
d. How long is the rock above a height of 17 m?
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The position function for a free-falling object is given by
h(t) = - 5t² + v0t + h0
where h = height in meters, t = time in secs., v0 = initial velocity in m./sec. and h0 = initial height in meters.
v0 = 12 m./sec.
h0 = 14 m.
Position function for this problem::
h(t) = - 5t² +12t + 14
a.) Initial Ht. = 14 m.
b.) Total Time: h(t) = 0:
- 5t² + 12t + 14 = 0
5t² - 12t - 14 = 0
5t² - 12t = 14
5(t² - 12/5 t) = 14
5(t² - 24/10 t) = 14
t² - 2.4 = 14 / 5
t² - 2.4 = 28/10
t² - 2.4 = 2.8
t² - 2.4 + 1.44 = 2.8 + 1.44
(t - 1.2)² = 4.24
t - 1.2 = √4.24
t - 1.2 = ± 2.06
t = 1.2 ± 2.06
If t = 1.2 + 2.06,
t = 3.26
If t = 1.2 - 2.06,
t = - 0.86
Since time cannot run backwards,
t = 3.26
The rock is in the air for 3.26 secs.
c.) h(t) > 17 m.
- 5t² + 12t +14 > 17
5t² - 12t - 14 > - 17
5t² - 12t > 14 - 17
5t² - 12t > - 3
5(t² - 12/5 t) > - 3
t² - 24/10 > - 3/5
t² - 2.4 > - 0.6
t² - 2.4 + 1.44 > - 0.6 + 1.44
(t - 1.2)² > 0.84
t - 1.2 > √0.84
t - 1.2 > ± 0.917
t > 1.2 ± 0.917
If t > 1.2 + 0.917,
t > 2.12
If t > 1.2 - 0.917,
t > 0.283
The rock is above 17 m. at 0.283 sec. rising and at 2.12 secs. falling.
d.) The rock is above 17 m. for 2.12 - 0.283 secs., or 1.837 secs.
h(t) = - 5t² + v0t + h0
where h = height in meters, t = time in secs., v0 = initial velocity in m./sec. and h0 = initial height in meters.
v0 = 12 m./sec.
h0 = 14 m.
Position function for this problem::
h(t) = - 5t² +12t + 14
a.) Initial Ht. = 14 m.
b.) Total Time: h(t) = 0:
- 5t² + 12t + 14 = 0
5t² - 12t - 14 = 0
5t² - 12t = 14
5(t² - 12/5 t) = 14
5(t² - 24/10 t) = 14
t² - 2.4 = 14 / 5
t² - 2.4 = 28/10
t² - 2.4 = 2.8
t² - 2.4 + 1.44 = 2.8 + 1.44
(t - 1.2)² = 4.24
t - 1.2 = √4.24
t - 1.2 = ± 2.06
t = 1.2 ± 2.06
If t = 1.2 + 2.06,
t = 3.26
If t = 1.2 - 2.06,
t = - 0.86
Since time cannot run backwards,
t = 3.26
The rock is in the air for 3.26 secs.
c.) h(t) > 17 m.
- 5t² + 12t +14 > 17
5t² - 12t - 14 > - 17
5t² - 12t > 14 - 17
5t² - 12t > - 3
5(t² - 12/5 t) > - 3
t² - 24/10 > - 3/5
t² - 2.4 > - 0.6
t² - 2.4 + 1.44 > - 0.6 + 1.44
(t - 1.2)² > 0.84
t - 1.2 > √0.84
t - 1.2 > ± 0.917
t > 1.2 ± 0.917
If t > 1.2 + 0.917,
t > 2.12
If t > 1.2 - 0.917,
t > 0.283
The rock is above 17 m. at 0.283 sec. rising and at 2.12 secs. falling.
d.) The rock is above 17 m. for 2.12 - 0.283 secs., or 1.837 secs.