Recent content by physics_swede

All my attempts to determine the optimal angle of launch with a vertical displacement of 1,5 have been unsuccessful. Is it impossible to calculate?

The initial velocity doesn't effect the optimal angle for a specific vertical displacement. When I use the formula above and plug in different values at v0 and 1,5 at yo I get different results. Logically, the equation for θ shouldn't involve v0 at all.

That is indeed a very interesting method. However, when I plug 1,5 at y0 I get something bigger than 1 under the square root. This means that the expression isn't defined since -1<cos^-1<1.

A vertical displacement does indeed change the optimal angle. Imagine that you're standing on a mountain 1000m above the ground level and are about to throw something. Due to the position the horizontal vector vcosθ needs to be much bigger than if the same situation would have occurred on the...

I'm trying to determine the angle that will result in the theoretical maximal range 1,5 meters above the x-axis in a coordinate system. Everyone knows that the optimal angle of launch is 45 degrees when launching from origo. A vertical displacement results in a slightly smaller angle.

I've been trying to determine the optimun angle of launch 1,5 metres above the ground by pooceeding from the basic equations for throwing motions: x=vcosθ*t y=y0+vsinθ*t-gt^2/2, y0=1,5m, g=9,8m/s^2 What I've done so far is expressing t as x/vcosθ. Insertion in the equation for y gives...