If we neglect air resistance, then the range of a ball

[Niaboc67]

Homework Statement

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 =

The Attempt at a Solution

If I understand correctly. To find the max and min you simply take the derivative of that function and set it to zero. In this case I assume we have only a max since a ball is being tossed into the air.

I get the functions derivative:
R'(θ) = V_0^2/g *2*cos(2θ)
Then isolate the function
cos(2θ) = 0, well that is only true for pi/2, since it's domain here is [0,pi/2]
therefore isolating θ by itself by dividing through.
2θ=pi/2

θ=pi/4

Is this process and reasoning correctly thought out?

Thank you
Os

 

[Mark44]

Homework Statement

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 =

The Attempt at a Solution

If I understand correctly. To find the max and min you simply take the derivative of that function and set it to zero. In this case I assume we have only a max since a ball is being tossed into the air.

I get the functions derivative:
R'(θ) = V_0^2/g *2*cos(2θ)
Then isolate the function
cos(2θ) = 0, well that is only true for pi/2, since it's domain here is [0,pi/2]
therefore isolating θ by itself by dividing through.
2θ=pi/2

θ=pi/4

Is this process and reasoning correctly thought out?

Thank you
Os

Looks good to me. For maximum range, the angle should be $\pi/4$.