Question 996070: If we neglect air resistance, then the range of a ball (or any projectile) shot at an angle θ with respect to the x axis and with an initial velocity v0, is given by
R(θ)=v_0^2/g sin(2θ) for 0 ≤ θ ≤ π/2
where g is the acceleration due to gravity (9.8 meters per second per second).
For what value of θ is the maximum range attained? (Note that the answer is numerical, not symbolic.)
θ =
Please explain
Thank you
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! R(theta) = (v0^2/g)*sin(2*theta)
R ' (theta) = (v0^2/g)*2*cos(2*theta)
0 = (v0^2/g)*2*cos(2*theta)
cos(2*theta) = 0
2*theta = arccos(0)
2*theta = pi/2 + 2pi*n or 2*theta = -pi/2 + 2pi*n
theta = pi/4 + pi*n or theta = -pi/4 + 2pi*n
Ignore the theta = -pi/4 + 2pi*n portion since the first part is negative pushing it out of the interval [0,pi/2]
Only one value of n will make theta = pi/4 + pi*n in the interval [0,pi/2]. This value of n is n = 0
If n = 0, then theta = pi/4
Final Answer: pi/4 = 0.78539816339744 (decimal value is approximate)
pi/4 radians = 45 degrees
This is halfway between 0 and 90 degrees
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