Derive Max Impact Height Equation?

Question

If given the trajectory equation as y=x*tan# - (g*x^2)/(2V^2*(cos#)^2)

where
y is the path of the trajectory
x is the horizontal distance
# is the angle it is launched at
g is gravity
V is launch velocity

how do you, using derivatives, find the equation to calculate the maximum height of an object being launch so that you can plug in velocity, gravity, and horizontal distance and equal the angle it should be launched at to reach maximum height

Answer 1

Find the first derivative dy/dx = tan# - 2gx/[2V^2 cos#^2] = 0 and solve for x = [2V^2 cos#^2]tan#/2g =v^2 cos# cos# sin#/cos#/g = V^2 cos# sin#/g; where #, V, and g are given initial launch conditions.

Then insert your x value from above into the original equation y = x*tan# - (g*x^2)/(2V^2*(cos#)^2); for the same launch conditions of #, V, and g. The y answer you get will be the maximum height, which will be reached when the missile is x distance down range.

This is not so much a physics problem as it is a math problem. The math you should know is that where dy/dx = 0 the slope of the trajectory is flat...that is, horizontal with the X axis. And that's the point where the missile levels off and, then, starts back down.

Answer 2

It outcomes from equating the kinetic ability gained in falling from relax in a continuing gravitational field of acceleration g to the potential ability lost falling by top h in that field so it applies whilst g does not replace very plenty over top h and whilst there at the instant are not any losses through friction, cavitation, etc. - the object should be at relax ( no ok.E.) to start with. v is the fee that the object is moving after it has moved by top h. - if it moved distance h+H initially yet then carried on up the pendulum swing or up a loop of song then the fee on the backside could be bigger than at h.