A long jumper leaves the ground at an angle of

22

°

above the horizontal, at a speed of

9

/

m

s

. The height of the jumper can be modeled by

=

h

x

+

−

0.046

x

2

0.362

x

, where

h

is the jumper's height in meters and

x

is the horizontal distance from the point of launch.

a) = 3.96

b) = 0.72

c) = 7.94

Step-by-step explanation:

Complete question.

(A long jumper leaves the ground at an angle of 20° above the horizontal, at a speed of 11 m/sec. The height of the jumper can be modeled by h(x) = -0.046x2+ 0.364x, 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.

Data

Θ = 20

v = 11m/s

h(x) = -0.046x² + 0.364x

x = horizontal distance

h = height

a.

h(x) = -0.046x² + 0.364x

at maximum height,

d / dx (h) = 2(-0.46x) + 0.364

d / dx (h) = -0.092x + 0.364

at maximum height d / dx (h) = 0

0 = -0.092x + 0.364

0.092x = 0.364

x = 3.96.

b).

Hmax = -0.046x² + 0.364x

Hmax = -0.046(3.96) + 0.364(3.96)

Hmax = -0.72 + 1.44

Hmax = 0.72

c) the length of the jump can be calculated using the formula for range of a projectile

R = v² sin2Θ / g

g = 9.8m/s²

R = 11² * sin 2(20) / 9.8

R = 121 * sin40 / 9.8

R = 77.777 / 9.8

R = 7.936 = 7.94