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.
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Answer:
Part a)
F = 0.735 N
Part b)

Explanation:
As we know that mass of the cube is given as




Now we know that cube is floating in the water
So net force due to weight of the cube must be counter balanced by buoyancy force on the liquid
so we have



Part b)
Percentage of volume submerged into the liquid is given as



now percentage of submerged liquid is given as


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