10. A military jet flies horizontally with a velocity of 450 m/s and 20,000 m above the ground. When the jet is straight above an artillery gun a shell is fired. Assuming the shell hits the jet.
a. Calculate the horizontal component of the initial velocity of the shell.
b. Calculate the vertical component of the initial velocity of the shell.
c. Calculate the magnitude of the initial velocity of the shell.
d. Calculate the angle of the initial velocity above the horizontal.
To answer these questions, we need to use the principles of projectile motion. Projectile motion occurs when an object is launched into the air and moves under the influence of gravity. In this case, the shell is fired horizontally from the artillery gun, while the jet is flying horizontally.
a. To calculate the horizontal component of the initial velocity of the shell, we can use the fact that the shell and the jet are moving horizontally at the same velocity, 450 m/s. Therefore, the horizontal component of the initial velocity of the shell is also 450 m/s.
b. To calculate the vertical component of the initial velocity of the shell, we need to consider the height at which the jet is flying above the ground. In this case, the jet is 20,000 m above the ground. The vertical component of the initial velocity of the shell can be determined using the equation for projectile motion:
v_fy = v_iy - gt
Where v_fy is the final vertical velocity, v_iy is the initial vertical velocity (which we want to find), g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time of flight. Since the jet and the shell are aligned vertically when the shell is fired, the time of flight can be determined using the equation:
d = v_it + 0.5gt²
Where d is the vertical distance between the shell and the jet (20,000 m), v_i is the initial vertical velocity, g is the acceleration due to gravity, and t is the time of flight (which we want to find).
Substituting the known values into the equation, we get:
20,000 = v_it + 0.5(9.8)t²
Simplifying, we have:
4.9t² + v_it - 20,000 = 0
This is a quadratic equation in terms of t. Solving for t, we get two solutions: t = 63.59 s and t = -63.59 s. Since time cannot be negative in this context, we can discard the negative solution. Therefore, the time of flight is approximately 63.59 s.
Now, we can substitute the value of t into the equation for v_iy:
v_iy = v_fy + gt
Since the shell and the jet will meet at the same height above the ground, the final vertical velocity of the shell (v_fy) will be 0 m/s. Plugging in the values, we have:
v_iy = 0 + (9.8)(63.59)
v_iy = 625.08 m/s
Therefore, the vertical component of the initial velocity of the shell is approximately 625.08 m/s.
c. To calculate the magnitude of the initial velocity of the shell, we can use the Pythagorean theorem. The magnitude of the initial velocity (V) is given by:
V = sqrt((Vx)² + (Vy)²)
Where (Vx) is the horizontal component of the initial velocity of the shell (450 m/s) and (Vy) is the vertical component of the initial velocity of the shell (625.08 m/s).
Plugging in the values, we have:
V = sqrt((450)² + (625.08)²)
V = sqrt(202,500 + 390,566.0064)
V = sqrt(593,066.0064)
V ≈ 769.81 m/s
Therefore, the magnitude of the initial velocity of the shell is approximately 769.81 m/s.
d. Finally, to calculate the angle of the initial velocity above the horizontal (θ), we can use the inverse tangent function. The angle θ can be determined using:
θ = arctan(Vy / Vx)
Plugging in the values, we have:
θ = arctan(625.08 / 450)
θ = arctan(1.38907)
θ ≈ 54.86°
Therefore, the angle of the initial velocity above the horizontal is approximately 54.86°.