Aspring-loaded gun can fire a projectile to a height h if it is fired straight up. if the same gun is pointed at an angle of 45◦ from the vertical, what maximum height h45◦ can now be reached by the projectile

When firing straight up:
v^2 = u^2 - 2gh, where v = final velocity = 0, u = initial velocity, g = gravitational acceleration, h = maximum height attained.

Then,
0 = u^2 - 2gh
u = Sqrt (2gh) (1)

When firing at 45°,
Initial velocity, U = u Sin 45 = Sqrt (2gh)·Sin 45

Maximum height, H = U^2*(Sin Ф)^2/2g

substituting;
H = [Sqrt (2gh)·Sin 45]^2*(Sin 45)^2]/2g
H = [2gh*(Sin 45)^2*(Sin 45)^2]/2g
H = [h*(Sin 45)^4] = h/4

Therefore, maximum height when the gun fires at 45° is a quarter of maximum height when the gun fires vertically up.

Ais the answer hope this