An artillery crew demonstrates its skill by firing a shell at an angle of 69 deg and then lowering the gun barrel and firing a second shell at a smaller angle of 22 deg in such a way that the two shells collide in midair. The initial velocity of the shells is 900 m/s. Find the time elapsed between the two firings. Neglect air resistance.
To find the time elapsed between the two firings, we can break down the problem into two separate components: the horizontal motion and the vertical motion.
First, let's analyze the horizontal motion of the shells. Since there is no air resistance, the horizontal motion is constant and the shells will travel at a constant velocity throughout their trajectory. This means that the horizontal distance covered by each shell will be the same.
Given that the initial velocity of the shells is 900 m/s and neglecting air resistance, the horizontal component of the velocity will remain constant. Let's call this velocity Vx.
Vx = V * cos(θ)
Where:
Vx - Horizontal component of the velocity
V - Initial velocity (900 m/s)
θ - Angle of the trajectory (69° for the first shell and 22° for the second shell)
For the first shell:
Vx₁ = 900 m/s * cos(69°)
For the second shell:
Vx₂ = 900 m/s * cos(22°)
Since the horizontal distances covered by the shells are the same, we can equate the two equations:
Vx₁ * t = Vx₂ * t
where t is the time elapsed between the two firings.
Now let's analyze the vertical motion of the shells. The vertical component of the velocity, Vy, changes over time due to gravity. However, since the two shells collide in midair, we can determine the time elapsed by equating their vertical displacements.
The vertical displacement of a projectile can be determined using the following formula:
Δy = Vy * t + (1/2) * g * t^2
Where:
Δy - Vertical displacement
Vy - Initial vertical component of the velocity
g - Acceleration due to gravity (approximately 9.8 m/s^2)
t - Time
By equating the two vertical displacements, we can solve for t.
Δy₁ = Δy₂
(Vy₁ * t) + (1/2) * g * t^2₁ = (Vy₂ * t) + (1/2) * g * t^2₂
Now we need to find the initial vertical component of the velocity for each shell, Vy₁ and Vy₂.
Vy = V * sin(θ)
For the first shell:
Vy₁ = 900 m/s * sin(69°)
For the second shell:
Vy₂ = 900 m/s * sin(22°)
Plug these values back into the equation and solve for t:
(Vy₁ * t) + (1/2) * g * t^2₁ = (Vy₂ * t) + (1/2) * g * t^2₂
Finally, solve for t to determine the time elapsed between the two firings.