Recent content by bluemyner

I'm not sure if I totally understand your first question but I'm taking taking the change in y as the equation that was given in the original problem. I figured find t would allow me to somehow find the change in x and then see where the trajectory intersect the parabola but I'm not sure how to...

I solved for t with the equation: 4(x)^1/2 = 0t + 1/2(9.8)(t)^2 t = (x)^1/4√(4/4.9) t = .904(x)^1/4 I'm not sure where I should/could go from here.

If I'm being completely honest, I have never done a problem where the projectile is being launched over a curved ground so I am not sure where that given equation plays a part in finding my x and y components.

I looked at the problem with a fresh set of eyes this morning. I resolved for time using the quadratic equation and got t = 2.91s and then solved for Final velocity in the y direction which gave me 24.56m/s. Using this and the initial velocity in the x direction and the Pythagorean theorem I...

Thanks for the welcome! I'll be completely honest, I am very lost with this problem and I'm new to Physics forums so I saw that it said to try and answer the problem and I gave it my best shot. I've been trying the plug and chug approach to this problem and it hasn't worked out for me.

Vx = 10cos(23) = 9.21m/s Vy = 10sin(23) = 3.91m/s -30 = 3.91t + 1/2(-9.8)t^2 0 = t(3.91 - 4.9t) + 30 4.9t = 33.91 t = 6.92s Delta X = 9.21(6.92) + 1/2(9.8)(6.92)^2 Delta X = 298.38m Vf^2 = 9.21 + 2(9.8)(298.38) Vf^2 = 5857.458 Vf = 76.53m/s