A long jumper leaves the ground at an angle of 20 o above the horizontal at a speed of11 m/sec. The height of the jumper can be modeled by h x = − 0.046 x 2 + 0.364 x , where h is the jumper’s height in meters and x is the horizontal distance from the point of launch. a. At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places. b. What is the maximum height of the long jumper? Round to 2 decimal places. c. What is the length of the jump? Round to 1 decimal place.
A long jumper leaves the ground at an angle of 20 o above the horizontal at a speed of11 m/sec. The height of the jumper can be modeled by h x = − 0.046 x 2 + 0.364 x , where h is the jumper’s height in meters and x is the horizontal distance from the point of launch. a. At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places. b. What is the maximum height of the long jumper? Round to 2 decimal places. c. What is the length of the jump? Round to 1 decimal place.
Solution Summary: The author calculates the horizontal distance from the point of launch at which the maximum height of the jumper occurs using f(x)=-0.046x2+0.364x.
A long jumper leaves the ground at an angle of
20
o
above the horizontal at a speed of11 m/sec. The height of the jumper can be modeled by
h
x
=
−
0.046
x
2
+
0.364
x
,
where h is the jumper’s height in meters and x is the horizontal distance from the point of launch.
a. At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places.
b. What is the maximum height of the long jumper? Round to 2 decimal places.
c. What is the length of the jump? Round to 1 decimal place.
T
1
7. Fill in the blanks to write the calculus problem that would result in the following integral (do
not evaluate the interval). Draw a graph representing the problem.
So
π/2
2 2πxcosx dx
Find the volume of the solid obtained when the region under the curve
on the interval
is rotated about the
axis.
38,189
5. Draw a detailed graph to and set up, but do not evaluate, an integral for the volume of the
solid obtained by rotating the region bounded by the curve: y = cos²x_for_ |x|
≤
and the curve y
y =
about the line
x =
=플
2
80
F3
a
FEB
9
2
7
0
MacBook Air
3
2
stv
DG
Find f(x) and g(x) such that h(x) = (fog)(x) and g(x) = 3 - 5x.
h(x) = (3 –5x)3 – 7(3 −5x)2 + 3(3 −5x) – 1
-
-
-
f(x) = ☐
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