University Physics 1 - Motion in 2D & 3DI need help with this problem and a very detailed explanation of the solutions described below:&1) using this - general equation of height (y) -> y = (V0 sina0 )t- 1/2 gt^2 2) eliminate t from y using first equation x=V0Cos(Degree)t 3) Plug in given values 4) equate with given equation to find x , then find y 48. A projectile is shot at a hill, the base of which is 300 m away. The projectile is shot at 60° above thehorizontal with an initial speed of 75 m/s. The hill can be approximated by a plane sloped at 20° to thehorizontal. Relative to the coordinate system shown in the following figure, the equation of this straight line isy = (tan20°)x - 109. Where on the hill does the projectile land?YA75 m/s60°y = (tan 20°)x - 109-300 m.20°

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Description: University Physics 1 - Motion in 2D & 3DI need help with this problem and a very detailed explanation of the solutions described below:&1) using this - general equation of height (y) -> y = (V0 sina0 )t- 1/2 gt^2 2) eliminate t from y using first

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Answered: 48. A projectile is shot at a hill, the…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Motion In Two Dimensions

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Problem 16P: A firefighter, a distance d from a burning building, directs a stream of water from a fire hose at...

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University Physics 1 - Motion in 2D & 3D
I need help with this problem and a very detailed explanation of the solutions described below:

1) using this - general equation of height (y) -> y = (V₀ sina₀)t- 1/2 gt^2

2) eliminate t from y using first equation x=V₀Cos(Degree)t

3) Plug in given values

4) equate with given equation to find x , then find y

48. A projectile is shot at a hill, the base of which is 300 m away. The projectile is shot at 60° above the
horizontal with an initial speed of 75 m/s. The hill can be approximated by a plane sloped at 20° to the
horizontal. Relative to the coordinate system shown in the following figure, the equation of this straight line is
y = (tan20°)x - 109. Where on the hill does the projectile land?
YA
75 m/s
60°
y = (tan 20°)x - 109
-300 m.
20°

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