Comments on: Hot air balloon physics problem!? http://diy-solarpanels.net/air-x-wind/hot-air-balloon-physics-problem Build your own solar panels for less then $200! Fri, 18 May 2012 11:27:59 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: John N http://diy-solarpanels.net/air-x-wind/hot-air-balloon-physics-problem/comment-page-1#comment-1472 John N Thu, 04 Mar 2010 07:20:59 +0000 http://diy-solarpanels.net/air-x-wind/hot-air-balloon-physics-problem#comment-1472 I assume that the balloon's initial velocity is the 17 m/s @ 32° above the horizontal. You need to take that and break it down into horizontal and vertical components. It might help to draw a right triangle with a hypotenuse of 17 and one angle of 32°. Label the horizontal leg "x" and the vertical leg "y." The vertical component can expressed as follows: sin 32° = y/17, and y = 9.01, so your initial vertical velocity is 9.01 m/s² upward. The horizontal component can be expressed as cos 32° = x/17, and x = 14.42, so your initial horizontal velocity is 14.42 m/s² horizontally. Now for your velocity equation: v = v(0) + at, where v(0) is your initial velocity a is your acceleration, and t is your time. You need to do this equation for the horizontal and vertical components. When you're done, draw another right triangle with one leg as your vertical velocity and one as your horizontal. Using the equation a² + b² = c², plug in your horizontal and veritcal velocities for a and b. Don't forget to take the square root of the sum for your resultant vector's magnitude. To find the angle, do the following tan θ = (vertical velocity / horizontal velocity), change to θ = tanˉ¹(vertical velocity / horizontal velocity), and take the inverse tangent of the vertical velocity divided by the horizontal velocity. That will give you the angle. Good luck. I hope this was helpful.<br><b>References : </b><br> I assume that the balloon’s initial velocity is the 17 m/s @ 32° above the horizontal. You need to take that and break it down into horizontal and vertical components. It might help to draw a right triangle with a hypotenuse of 17 and one angle of 32°. Label the horizontal leg "x" and the vertical leg "y."

The vertical component can expressed as follows:

sin 32° = y/17, and y = 9.01, so your initial vertical velocity is 9.01 m/s² upward.

The horizontal component can be expressed as

cos 32° = x/17, and x = 14.42, so your initial horizontal velocity is 14.42 m/s² horizontally.

Now for your velocity equation:

v = v(0) + at, where v(0) is your initial velocity a is your acceleration, and t is your time.

You need to do this equation for the horizontal and vertical components. When you’re done, draw another right triangle with one leg as your vertical velocity and one as your horizontal. Using the equation a² + b² = c², plug in your horizontal and veritcal velocities for a and b. Don’t forget to take the square root of the sum for your resultant vector’s magnitude.

To find the angle, do the following

tan θ = (vertical velocity / horizontal velocity), change to

θ = tanˉ¹(vertical velocity / horizontal velocity),

and take the inverse tangent of the vertical velocity divided by the horizontal velocity.

That will give you the angle. Good luck. I hope this was helpful.
References :

]]> By: Alexey V http://diy-solarpanels.net/air-x-wind/hot-air-balloon-physics-problem/comment-page-1#comment-1471 Alexey V Thu, 04 Mar 2010 06:42:59 +0000 http://diy-solarpanels.net/air-x-wind/hot-air-balloon-physics-problem#comment-1471 Initial y component is Vy(0) = 17sin(32), so it will change as Vy(t) = 17sin(32) + 3.4t. For x component similarly Vx(t) = 17cos(32) + 2.5t. Substitute your t = 45 seconds and find total velocity from Pythagorean theorem V = sqrt(Vx^2 + Vy^2)<br><b>References : </b><br> Initial y component is Vy(0) = 17sin(32), so it will change as Vy(t) = 17sin(32) + 3.4t. For x component similarly Vx(t) = 17cos(32) + 2.5t. Substitute your t = 45 seconds and find total velocity from Pythagorean theorem V = sqrt(Vx^2 + Vy^2)
References :

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