Suppose a projectile is launched with velocity v at an angle x to the horizontal from the base of a hill that makes an angle a with the horizontal (x>a). Then the range of the projectile, measured along the slope of the hill, is given by R= 2v^2cosxsin(x-a)/gcos^2a. Show that if a=45, then R=v^2(sqrt2)/g * (sin2x-cos2x-1).
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Trigonometry Problem using Double and Half Identities
- Thread starter mjonascu
- Start date
D
Deleted member 4993
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Suppose a projectile is launched with velocity v at an angle x to the horizontal from the base of a hill that makes an angle a with the horizontal (x>a). Then the range of the projectile, measured along the slope of the hill, is given by
R= 2v^2cos(x)*sin(x-a)/g * cos^2(a).
Show that if a=45°,
then
R=v^2(sqrt2)/g * [sin(2x)-cos(2x)-1].
Does your problem look like above?
In any case replace 'a' by 45° and continue...
Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
D
Deleted member 4993
Guest
I did that and I got R=2v^2cosxsin(x-sqrt2/2)/(g/2). I still am not sure how to get it to equal to v^2(sqrt2)/g * (sin2x-cos2x-1).
That is incorrect.
R=2v^2cos(x)*sin(x-45)/(g/2)
= 4/g * v^2 * cos(x) * [sin(x)*cos(45°) - sin(45°)*cos(x)]
= 4/g * v^2 * cos(x) * [sin(x)*1/√2 - 1/√2 * cos(x)]
= 4/g * 1/√2 * v^2 * cos(x) * [sin(x) - cos(x)]
Now continue....