A hot air balloon is rising at a rate of 3.4 m/s^2and a horizontal wind is moving it at a rate of 2.5 m/s^2. If the balloon is moving at 17 m/s @ 32o above the positive x axis, what is the final velocity after 45 seconds?
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.
March 4th, 2010 at 1:42 am
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 :
March 4th, 2010 at 2:20 am
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 :