When the gun's angle of elevation was set to 45°, what would be the range?

Question

The citizens of Paris were terrified during World War I when they were suddenly bombarded with shells fired from a long-range gun known as Big Bertha. The barrel of the gun was 36.6 m long and it had a muzzle speed of 1.41 km/s. When the gun's angle of elevation was set to 45°, what would be the range? For the purposes of solving this problem, neglect air resistance.

Answer 1

Let V= initial velocity at 45 deg, Vv= vertical velocity, Vh=horizontal velocity, t1=upward time, t2= downward time, h1= vertical height, and R=horizontal distance. Then,

Vv=1.41*Sine 45
Vv=1km/s
Vh=1.41*cosine 45
Vh=1 km/s
h1=Vv^2/(2*g)
h1=(1000m)^2/(2*9.81)
h1=50968.4 meters=50.9684 km (ideal!)
t1=Vv/g
t1=1000/9.81
t1=101.9 seconds
t2=t1=101.9 seconds
R=Vh*(t1+t2)
R=1*(101.9+101.9)
R=203.8 km (ideal!)

Answer 2

I love it when they make the number crunching easy.

1.41 at an angle of 45 degrees means you have a vertical velocity of 1 km/s and a horizontal velocity of 1 km/s .

Vertical velocity 1.41 x sin (45) = 1
Horizontial velocity 1.41 cos (45) = 1

First calculate the time of flight by looking only at the vertical motion.

Height = Vt - (g/2)t^2 solve for t at height = zero you're looking for the time for the shell to travel up and come back down. So t > 0. Make sure the units used for g (gravitational acceleration) are consistent with units used for velocity Km or m.

Use that time to determine the Horizontal distance, since we're ignoring air friction there is no acceleration term and the formula is Distance = Vt

Answer 3

all i know is 45 degrees is the angle to achieve maxium distance

Answer 4

1.3 miles