# Thread: trig identitities applied question

1. ## trig identitities applied question

An object is propelled upward at an angle theta, 45 degree < theta <90 degree, to the horizontal with an initial velocity of v0 feet per second from the base of an inclined plane that makes an angle of 45 degree with the horizontal. If air is ignored, the distance R that it travels up the inclined plane is given by:
R= v0square square root of 2/16 multiplied by [cos(theta)(sin(theta)-cos(theta)]....Show R(THETA)

i really dont know how to solve this....this is really advanced and i need help with this

2. ## Re: trig identitities applied question

Originally Posted by koolaid123
An object is propelled upward at an angle theta, 45 degree < theta <90 degree, to the horizontal with an initial velocity of v0 feet per second from the base of an inclined plane that makes an angle of 45 degree with the horizontal. If air is ignored, the distance R that it travels up the inclined plane is given by:
R= v0square square root of 2/16 multiplied by [cos(theta)(sin(theta)-cos(theta)]....Show R(THETA)
show what?

3. ## Re: trig identitities applied question

Originally Posted by koolaid123
R= v0square square root of 2/16 multiplied by [cos(theta)(sin(theta)-cos(theta)]....Show R(THETA)
For clarification, you are being asked to prove the above formula?
$\displaystyle R= v_0^2 \sqrt {2/16} \cdot [cos(\theta) \left ( sin(\theta)-cos(\theta) \right ) ]$

This is confusing if R is a range. The units are all wrong. Please give us more information.

-Dan

PS Nope. Either R(theta) isn't the range or your answer is incorrect.

4. ## Re: trig identitities applied question

R= v0square square root of 2/16 multiplied by [cos(theta)(sin(theta)-cos(theta)]
assuming $\displaystyle R$ is the distance from launch to impact up the 45 degree incline ...

$\displaystyle \Delta x = v_0 \cos{\theta} \cdot t = \frac{R\sqrt{2}}{2}$

$\displaystyle t = \frac{R\sqrt{2}}{2v_0\cos{\theta}}$

$\displaystyle \Delta y = v_0 \sin{\theta} \cdot t - 16t^2 = \frac{R\sqrt{2}}{2}$

$\displaystyle v_0 \sin{\theta} \cdot \frac{R\sqrt{2}}{2v_0\cos{\theta}} - 16\left(\frac{R\sqrt{2}}{2v_0\cos{\theta}}\right)^ 2 = \frac{R\sqrt{2}}{2}$

$\displaystyle \tan{\theta} - \frac{8R\sqrt{2}}{v_0^2\cos^2{\theta}} = 1$

$\displaystyle R = \frac{v_0^2 \sqrt{2} \cdot \cos^2{\theta}(\tan{\theta} - 1)}{16}$

$\displaystyle R = \frac{v_0^2\sqrt{2} \cdot (\cos{\theta}\sin{\theta} - \cos^2{\theta})}{16}$

5. ## Re: trig identitities applied question

Oops! I solved for x, not R.

And the units are part of the 16. (I typically leave this as g/2.)

Thanks skeeter!

-Dan