Okay I have derived both equations.
dR/dx = v*(cos(x)t' - sin(x)t)
dH/dx = v*(cos(x)t + sin(x)t') - gt*t'
I am unsure where to take these results. Can you help me?
I'm not that familiar with this strategy of differentiation. But I thank you for giving me this new technique to me. I definitely have to try this out, and hopefully master it.
Thanks for your replies. I did that. But I got the same displacement equation that I posted.
t = (vsin(x)+sqrt(v^2*sin^2(x)+2gh))/g. I got this for the time period.
I came up with this problem and did the work, that is hardly an example of low-effort. I don't think I understand the hint well. Do you happen to know the solution?
Sorry about the messy equations. When I took the derivative of range vs angle, I got a very complicated output. Therefore I thought it was difficult to optimize using that solution to find an expression for the angle. That's why I thought there may be a different or better approach available.
I found the function of the range of the projectile launched from a cliff.
R = vcos(x)((vsin(x))+sqrt(v^2sin^2(x)+2gh))/g
I stopped here because I feel like taking the derivative and optimizing for maximum would spiral out of control.
Is there another approach to this problem?
For this problem I tried to find when the binoculars reaches the maximum height. So, (0.75 + 1.28)/2 = 1.015s. Using that information I can solve for the initial velocity. v_o = gt = 9.947 m/s.
Then using the initial velocity I can solve for the height of John using the 2nd kinematics equation...