A shell is fired from the ground with an initial speed of 1.54 103 m/s at an initial angle of 46° to the horizontal.
(a) Neglecting air resistance, find the shell's horizontal range.
R = Vo^2*sin(2A)/g.
R = (1540)^2*sin(92)/9.8 = 241,853 m.
To find the horizontal range of the shell, we can use the equation for horizontal range:
R = (v^2 × sin(2θ)) / g
where:
R is the range,
v is the initial velocity of the shell (1.54 × 10^3 m/s),
θ is the angle of elevation (46°), and
g is the acceleration due to gravity (9.8 m/s^2).
Substituting these values into the equation, we have:
R = (1.54 × 10^3 m/s)^2 × sin(2 × 46°) / 9.8 m/s^2
To calculate this value, we can use a calculator:
1. Calculate the value of sin(2 × 46°):
sin(92°) ≈ 0.999848
2. Square the initial velocity:
(1.54 × 10^3 m/s)^2 ≈ 2371 × 10^3 m^2/s^2 ≈ 2.371 × 10^6 m^2/s^2
3. Substitute the values into the equation and solve:
R = (2.371 × 10^6 m^2/s^2) × 0.999848 / 9.8 m/s^2
R ≈ 242,188.6 m
Therefore, neglecting air resistance, the shell's horizontal range is approximately 242,188.6 meters.