Recent content by DeltaForce

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.

This is an amazing solution. Thanks for taking the time to write this. The source you linked also looked fantastic. Thanks!

Like R = vcos(x)t and H = vsin(x)t - 1/2gt^2?

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?

Can you show me how you would do this problem? You seem like you know what's up.

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?

Hahahaha, I was sitting there wringing my brain out figuring out what I did wrong. Now I see. Thank you!

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...

Thanks, that's what I was looking for.

Question.

Yeah. I got it with the trig identities.

Ohh... ok. So it has something to do with the sum and difference identities. Thank you.