Could someone please help me with this question?
1 answer:
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Answer:
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
Answer:
Angular velocity (w) = 8.86 rad/s
Explanation:
Angular velocity (w) =
g= 9.81 m/s
R= 0.5
hi (initial depth) = 0.5m
Hence= = 8.86 rad/s