trig identitities applied question

**koolaid123** · November 19th 2012, 02:25 PM

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

**skeeter** · November 19th 2012, 02:36 PM

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?

**topsquark** · November 19th 2012, 03:03 PM

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?
$R= v_0^2 \sqrt {2/16} \cdot [cos(\theta) \left ( sin(\theta)-cos(\theta) \right ) ]$

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.

**skeeter** · November 19th 2012, 04:33 PM

R= v0square square root of 2/16 multiplied by [cos(theta)(sin(theta)-cos(theta)]

assuming $R$ is the distance from launch to impact up the 45 degree incline ...

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

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

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

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

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

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

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

**topsquark** · November 20th 2012, 06:37 AM

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

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