Good day to you In this particular problem in textbook, essential university physics by richard wolfson volume 1 4th edition. I don't understand how the intersection angle was determined (the 15 degrees one). Much appreciated

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Good day to you

In this particular problem in textbook, essential university physics by richard wolfson volume 1 4th edition. I don't understand how the intersection angle was determined (the 15 degrees one).

Much appreciated

< Chapter 3, Problem 85P
Given information:
The player kicks the ball 28 m on a level ground with its initial
velocity at 40° to the horizontal.
Calculation:
The trajectory intercepts a 15° slope.
Sketch the path of the trajectory and the slope as shown in Figure 1.
Trajectory
Slope
15°
Figure 1
Let the initial velocity be vo and the initial angle be 0.
The value of 0 = 40°.
The trajectory of the ball intersects an angle of 01
= 15° is
expressed as, y = x tan (01).
The expression for horizontal distance the player kicks the ball
with a slope of 15° (x) is shown below:
tan(6)–tan(4)) (1)
x = 2vcos? (0)
X =
Here, gis the acceleration due to gravity.
The ball on the level ground is xf = 28 m.
The init
Transcribed Image Text:< Chapter 3, Problem 85P Given information: The player kicks the ball 28 m on a level ground with its initial velocity at 40° to the horizontal. Calculation: The trajectory intercepts a 15° slope. Sketch the path of the trajectory and the slope as shown in Figure 1. Trajectory Slope 15° Figure 1 Let the initial velocity be vo and the initial angle be 0. The value of 0 = 40°. The trajectory of the ball intersects an angle of 01 = 15° is expressed as, y = x tan (01). The expression for horizontal distance the player kicks the ball with a slope of 15° (x) is shown below: tan(6)–tan(4)) (1) x = 2vcos? (0) X = Here, gis the acceleration due to gravity. The ball on the level ground is xf = 28 m. The init
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