During world war i, the germans had a gun called big bertha that was used to shell paris. the shell had an initial speed of 2.61 km/s at an initial inclination of 81.9° to the horizontal. the acceleration of gravity is 9.8 m/s^2. how far away did the shell hit? answer in units of km how long was it in the air? answer in units of s.

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

did yall get answers for this yet