glx) = -0.04x2 + 2.1x + 6.1 Shot released at 65° 40 Maximum height 30 Maximum height Distance of throw or maximum horizontal distance 10 flx) = -0.01x2 + 0.7x + 6.1 Shot released at 35° 10 20 30 40 50 60 70 80 90 Shot Put's Horizontal Distance (feet) Shot Put's Height (feet) 20

Glencoe Physics: Principles and Problems, Student Edition
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ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter6: Motion In Two Dimensions
Section: Chapter Questions
Problem 84A
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An athlete whose event is the shot put releases the shot with the same
initial velocity but at different angles. The figure shows the parabolic paths for shots released at angles of 35° and 65°. Solve the following based on the functions that model the parabolic paths.

When the shot whose path is shown by the red graph is released at an angle of 65°, its height, g(x), in feet, can be modeled by g(x) = -0.04x2 + 2.1x + 6.1, where x is the shot’s horizontal distance, in feet, from its point of release. Use this model to solve the below parts (a) through (c) and verify your answers using the red graph.

a. What is the maximum height, to the nearest tenth of a foot, of the shot and how far from its point of release does this occur?

b. What is the shot’s maximum horizontal distance, to the nearest tenth of a foot, or the distance of the throw?

c. From what height was the shot released?

glx) = -0.04x2 + 2.1x + 6.1
Shot released at 65°
40
Maximum height
30
Maximum height
Distance of throw or maximum
horizontal distance
10
flx) = -0.01x2 + 0.7x + 6.1
Shot released at 35°
10
20
30
40
50
60
70
80
90
Shot Put's Horizontal Distance (feet)
Shot Put's Height (feet)
20
Transcribed Image Text:glx) = -0.04x2 + 2.1x + 6.1 Shot released at 65° 40 Maximum height 30 Maximum height Distance of throw or maximum horizontal distance 10 flx) = -0.01x2 + 0.7x + 6.1 Shot released at 35° 10 20 30 40 50 60 70 80 90 Shot Put's Horizontal Distance (feet) Shot Put's Height (feet) 20
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