Answer:
a) = 3.96
b) = 0.72
c) = 7.94
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