Answer:
The shell hit at a horizontal distance 149.41 km
Explanation:
The shell had an initial speed of 1.4 km/s
Initial speed = 1400 m/s
Inclination = 65.8° to the horizontal
First let us find the time of travel
Consider the vertical motion,
We have equation of motion, s = ut + 0.5 at²
Initial velocity, u = 1400 sin 65.8 = 1276.97 m/s
Acceleration, a = -9.81 m/s²
Displacement, s = 0 m
Substituting
s = ut + 0.5 at²
0 = 1276.97 x t + 0.5 x (-9.81) xt²
t = 260.34 s
Now let us find the horizontal displacement,
Consider the horizontal motion,
We have equation of motion, s = ut + 0.5 at²
Initial velocity, u = 1400 cos 65.8 = 573.89 m/s
Acceleration, a = 0 m/s²
Time, t = 260.34 s
Substituting
s = ut + 0.5 at²
s = 573.89 x 260.34 + 0.5 x 0 x 260.34²
s = 149407.11 m = 149.41 km
The shell hit at a horizontal distance 149.41 km
