real hard calculus problem innvoles phisics! ANYONE?

Question

If we neglect air resistance, then the range of a ball (or any projectile) shot at an angle theta with respect to the x axis and with an initial velocity v, is given by

R(theta)=(v^2)/g sin(2 theta) text

in [0 , Pi/2]

where g is the acceleration due to gravity (9.8 meters per second per second).


For what value of theta is the maximum range attained?

Answer 1

We take the derivative of R with respect to theta.

v^2/g can be treated as a constant.

dR/dtheta = v^2/g * cos(2 theta) * 2



Set it equal to 0 to find the maximum.

v^2/g * cos(2 theta) * 2 = 0

cos(2 theta) = 0

2 theta = arccos(0)

2 theta = pi/2

theta = pi/4

Answer 2

To find a local maximum or minimum, take the derivative, and set it equal to 0. Then solve for the unknown.



In this case, R is what we want to maximize, and theta is the unknown.



dR / d theta = (v^2)/g 2 cos (2 theta) = 0



v and g aren't 0, they're constants. cos (2 theta) =0.



cos (beta) = 0 when beta = 90, -90 deg (or +/- pi / 2),

so if beta= 2 theta, then theta = beta / 2 = 45 degrees.



If you thought this physics was hard, try 3rd year university electromagnetics - calculus in 3 dimensions using cylindrical coordinates!

Answer 3

greetings



dR/dtheta = (v^2)/g * 2 *cos(2 theta) = 0 for min/max theta.



v and g are constants so cos(2 theta) = 0



on [0 , pi/2] cos is 0 at pi/2 so 2*theta = pi/2 and theta = pi/4



Now the 2nd derivative test...



d2R/dtheta2 = -4(v^2)/g*sin(2theta) < 0 because sin (2 theta) = sin (pi/2) =1



so theta = pi/4 is a maximum for the range.



Regards

Answer 4

45 degrees. The sine function is maximal (1.00) at 90 degrees, and the sine of twice an angle must be maximal at 45 degrees. The calculus is trivial: d/dx (sin(x)) = cos (x); cos (x) = 0 at x = 90 degrees.

Answer 5

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