solve for h in terms of Vy and t. Physics math. can anyone help?

Question

Vyt - gt^2/2 = h - gt^2/2
simplify the expression above and slove for h in terms of Vy and t
t=d/Vy substitute this value for t into your simplified equation.
Vx= v cos(angle) and Vy= sin(angle) subtitute these values for Vx and Vy in your current equation and simplify. divide both sides by d and you should have aand expression equivalent to tan(angle)= h/d
so how does all this work. I want explanations.

Answer 1

Simplify the equation to get
Vy*t=h

substitute in
t=d/Vy to get
d=h?

Not exactly sure what you're asking.
I think you meant that

h=Vy*t - g*t^2/2

which is the equation of motion for an object launched from y=0 at t=0 with a speed in the vertical of Vy and h is the height of apogee.

When the object reaches apogee, v(t)=0, where
the general v(t)=Vy-g*t
so t at apogee is
t=Vy/g
so
h=Vy^2/g-Vy^2/(2*g)
or
h=Vy^2/(2*g)

The range is found from
x(t)=Vx*t
when y(t)=0
or
t=2*Vy/g
which is
d=2*Vx*Vy/g

If the initial velocity magnitude is v, and the angle of launch trajectory is th, then
Vy=v*sin(th)
Vx=v*cos(th)

so h/d=(Vy^2/(2*g))/(2*Vx*Vy/g)
simplify
h/d=.25*Vy/Vx
h/d=tan(th)/4

j

Answer 2

as we are fixing for h, we prefer each and all of the words with h on the LHS consequently we are saying: y = sh/(s+h) y(s+h)=sh ys+yh =sh ys = sh - yh sh - yh =ys h(s-y) = ys h = ys/ (s-y) because of the fact we've all h words on one area, this is the superb answer