what max height can now be reached?

Question

A spring-loaded gun can firre a projectile to a
height, h, if it is fired straight up.

If the same gun is pointed at an angle of
45 degrees from the vertical, what maximum height
h45degrees can now be reached by the projectile?

1. h45degrees = h
2. h45degrees = h/2*sqrt(2)
3. h45degrees = h /4
4. h45degrees = h / 2
5. h45degrees = h / sqrt(2)

Answer 1

It is 4 - h/2
simple.
ok now we use this formula
vf^2 = vi^2 + 2ad
we will see the fires at velocity v and vf will be zero at the top, so:
0 = v^2 + 2 * a* h
we solve for v
it gives
v = sqt(2*10*h)
v = sqt(20h)
now when the angle is 45 degree
we use sin(45) * v to find the vertical vector if velocity since it determines the height.
so ...
sqt(2)/2 * sqt(20h) = v
v = sqt(40h)/2
no we use the same formula as we used earlier
0 = (sqt(40h/2))^2 + 2 * a * x
x is the new height
when we solve for x
we will get h/2
vola

Answer 2

v^2 - 0 = 2ah
h = (v^2)/2a
h2 = ((v√2)/2)^2/2a
h2 = ((2v^2)/4)/2a
h2 = (2/4)(v^2)/2a
h2 = (1/2)h