How can I get to ymax = v^2sin^2θ / 2g from y=xtanθ-(gx^2)/(2v^2 cos^2⁡θ )

Question

Recently I've been working on something in which I found the equation y=xtanθ-(gx^2)/(2v^2 cos^2⁡θ ) (equation for projectile motion), however, now I'm looking to find the maximum height equation which is ymax = v^2sin^2θ / 2g, but I can't seem to make the first equation become the second equation. Any help?

Answer 1

The easiest way to solve this problem is as follows:

You find the flight time using the equation: y=v sinθ t -1/2 g t^2, which gives you t(flight)= ( 2 v sinθ ) / g. This is when y=0. Due to the symmetry of your problem, i.e your initial position and your final position are on the same horizontal line, half of the flight time gives you the time for the maximum height. Then, t(max)= (v sinθ) / g. Inserting this into the y equation: y(max)= v sinθ (v sinθ) / g - 1/2 g ((v sinθ) / g)^2 gives you the desired result: y(max)= (v^2 (sinθ)^2 ) / (2g)

There is also another way to solve this using calculus. Have you learned calculus yet?