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


by Shadow37
Tags: angle, height, maximum, projectile, speed
Shadow37
Shadow37 is offline
#1
Apr10-12, 07:49 PM
P: 5
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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azizlwl
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#2
Apr10-12, 10:10 PM
P: 961
Hope you can show all the equations used in your workings.
Shadow37
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#3
Apr10-12, 11:27 PM
P: 5
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?

Shadow37
Shadow37 is offline
#4
Apr10-12, 11:39 PM
P: 5

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


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.
Genoseeker
Genoseeker is offline
#5
Apr11-12, 12:00 AM
P: 10
Quote Quote by Shadow37 View Post
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..
Shadow37
Shadow37 is offline
#6
Apr11-12, 12:31 AM
P: 5
Quote Quote by Genoseeker View Post
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?
Genoseeker
Genoseeker is offline
#7
Apr11-12, 12:41 AM
P: 10
Quote Quote by Shadow37 View Post
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
Shadow37
Shadow37 is offline
#8
Apr11-12, 12:46 AM
P: 5
Thanks for replying.
Genoseeker
Genoseeker is offline
#9
Apr11-12, 02:28 AM
P: 10
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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