This problem is going to be pretty long to solve. So, prepare.
We’re interested in the change in our x position. So we have to break the velocity vector up into its components. Do cosine of 50 and then multiply by the magnitude of the velocity. I got 20.57m/s. That’s our initial velocity. And remember, horizontal acceleration is zero. The vertical acceleration, or any vertical component, has no effect on the horizontal components. In order to solve this problem, we want to utilize this equation: Change in x-position = Vix*t
Let’s solve for time, which is dependent on the vertical components. The projectile will stop when it vertically hits the ground. Generally you want to use this equation for solving for time: Yf = Yi + Viy*t + 1/2at^2 We didn’t solve for the vertical component yet, so let’s do that now. (Sine of 50)*(32) = 24.51m/s Let’s now plug everything in: 0 = 0 + 24.5t - 4.9t^2 0 = 24.5t - 4.9t^2 0 = t(24.5 - 4.9t) -24.5 = -4.9t t = 5 seconds
The hard stuff is pretty much over. Put that 5 seconds into the other equation I said we wanted to use to solve the problem Change in x-position (range) = (20.57)*(5) = 102.85 meters
