A cannon ball is fired with a muzzle velocity of 300.00 m/sec at an angle of 45⁰ (cont.)?

Take g = 9.8 m/s²
Take all upward quantities to be positive.

a.
Consider the vertical component of the motion (uniform acceleration motion):
s(y) = u(y) t + (1/2) a(y) t²
0 = (300 sin45°) t + (1/2) (-9.8) t²
t ≠ 0, and thus t = 43.3 s

Consider the horizontal component of the motion (uniform velocity):
Maximum horizontal range = u(x) t = (300 cos45°) × 43.3 = 9190 m

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b.
Consider the vertical component of the motion:
s(y) = u(y) t + (1/2) a t²
s(y) = (300 sin45°) (1.20) + (1/2) (-9.8) (1.20)² = 248 m

Consider the horizontal component of the motion:
s(x) = u(x) t = (300 cos45°) (1.20) = 255 m

Distance between the final and initial positions = √(248² + 255²) = 356 m
Angle made between the final and initial positions = tan⁻¹(248/255) = 44.2°

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c.
Consider the vertical component of the motion:
v(y) = u(y) + a(y) t
v(y) = (300 sin45°) + (-9.8) (2.00) = 193 m/s

Magnitude of velocity = √[193² + (300 cos45°)°] = 287 m/s
At an angle made with horizontal = tan⁻¹[193/(300 cos45°)] = 42.3°

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d.
Consider the vertical component of the motion:
v(y)² = u(y)² + 2 a(y) s(y)
0 = (300 sin45°)² + 2 (-9.8) s
Height it will going up, s(y) = 2300 m

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e)
Consider the vertical component of the motion:
s(y) = u(y) t + (1/2) a(y) t
0 = (300 sin45°)t + (1/2) (-9.81) t²
t ≠ 0, and thus time taken when it stays in the air, t = 43.2 s

Θ o = 45°, therefore :
Vox = Vo*cos 45° = 300*√ 2 / 2 = 150√ 2 m/sec (≅ 212)
Voy  = Vo*sin 45° = 300*√ 2 / 2 = 150√ 2 m/sec (≅ 212)

a. What is its maximum horizontal range R ?
R = Vo^2/(1000*g)*sin (2Θ o) = 300^2/9,807*1 = 9.18 km

b. Where is the ball after 1.20s?
Y = Voy*1.2-g/2*1.2^2 = 180√ 2-4.9035*1.44 = 247 m
X = Vox*1.2 = 180√ 2 m (≅ 255)

c. What is its velocity V after 2.00s?
Vy = Voy-g*t = 150√ 2-9,807*2 = 193 m/sec
Vx = Vox = 212 m/sec
V = √ 212^2+ 193^2 =  286 m/sec 

d. At what height hmax will it go upward?
hmax = Voy^2/2g = 150^2*2/19.614 = 2295 m (just R/4) 

e. How long t will it stay in the air?
t =2*tup = 2*Voy/g = 30.6√ 2 sec (≅ 43.3)