Calculate the speed of a projectile from the angle and maximum height.

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Homework Help Overview

The problem involves calculating the horizontal speed of water from a fountain that sprays at a 50-degree angle, reaching a maximum height of 0.15 meters. Participants are exploring the relationships between the angles, heights, and velocities involved in projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of trigonometric functions, specifically SOHCAHTOA, to relate the height of the arc to the horizontal distance. There are questions about how to connect the calculated distances to the speed of the water. Some participants express uncertainty about the application of kinematic equations and their relevance to the problem.

Discussion Status

The discussion is ongoing, with participants sharing their attempts to apply various equations and concepts. Some guidance has been offered regarding the use of kinematic equations, but there is no explicit consensus on the next steps or the correct application of the formulas.

Contextual Notes

Participants mention constraints such as the need to show all equations used and the challenge of connecting theoretical understanding with practical application. There is also a repeated emphasis on not wanting to appear lazy, indicating a struggle with the problem despite significant effort.

Shadow37
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The problem goes like this. A water fountain sprays water at a 50 degree angle from the horizontal. The top of the arc is .15m high. What is the horizontal speed of the water?

Using SOHCAHTOA, I have managed to find the distance the water flies, and the angles of the triangle the 50 degree angle forms, but that is it. Can anyone help?
 
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Hope you can show all the equations used in your workings.
 
The use of SOHCAHTOA is permitted because our triangle has a 90° angle from the height of the fountain to the top of the arc. The θ is 50°, and the third angle is 40°.

From SOHCAHTOA, I used the TOA portion. Tangent of θ = opposite over adjacent.

tanθ = .15m / adjacent. Since θ is 50°, you can solve for adjacent, which is equal to -.552m.

This is the distance from the 50° to the 90°, in other words, from the fountain to the place below the top of the arc. Doubling this, you get the distance from the fountain to the place the water lands.

How do I go from here to the speed of the water?
 
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Please, do not take me as being lazy. I have sat on this problem for hours now, and it still does not work for me. I have looked online, and went through several chapters in my physics book. I am pretty sure the answer is quite simple, but for some foolish reason, I cannot connect the dots.
 
Shadow37 said:
Please, do not take me as being lazy. I have sat on this problem for hours now, and it still does not work for me. I have looked online, and went through several chapters in my physics book. I am pretty sure the answer is quite simple, but for some foolish reason, I cannot connect the dots.

Vy = squareroot (2gh)

from the formula Vf^2 = Vi^2 +2*a*y
Vx = Vy / tan (theta)

TOA..
 
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Genoseeker said:
Vy = squareroot (2gh)

from the formula Vf^2 = Vi^2 +2*a*y I have this formula, but not the accelaration.
Vx = Vy / tan (theta) What is this portion for?

TOA..
Hold on a sec. Let me get this straight, vertical velocity = squareroot (2 x 9.8 x .15) Yes?
 
Shadow37 said:
Hold on a sec. Let me get this straight, vertical velocity = squareroot (2 x 9.8 x .15) Yes?

yep that is the simplified version of one of the 5 kinematics equations.

and is ONLY true for free falling objects in parabolic motion.

Vyi = squareroot (2 * 9.8 * MAX Height)

it does not work all the time but i does for your case
 
Thanks for replying.
 
Vx = Vy / tan (theta) What is this portion for?

Answer. they are asking for the horizontal velocity and this finds the horizontal velocity from the vertical velocity.
 

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