A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an initial angle of 43.0◦ to the horizontal.
The acceleration of gravity is 9.81 m/s2 .
a) Neglecting air resistance, find the shell’s horizontal range.
Answer in units of m.
b) How long is the shell in motion? Answer in units of s.
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a) Well, if we neglect air resistance, that means the shell will be as free as a bird. A dangerous, explosive bird, but still a bird. Now, let's calculate the horizontal range.
To do this, we need to split the initial velocity into its horizontal and vertical components. The horizontal component of the velocity will remain constant throughout the motion, while the vertical component will change due to gravity.
The initial velocity can be broken down into its components as follows:
Vx = V₀ * cos(θ)
Vy = V₀ * sin(θ)
where V₀ is the initial speed and θ is the launch angle.
Using these equations, we find:
Vx = 1.60 × 10³ m/s * cos(43.0°)
Vx = 1.60 × 10³ m/s * 0.7314
Vx ≈ 1165.12 m/s
Now, we can calculate the time it takes for the shell to reach maximum height. The vertical velocity component will be zero at this point. Using the equation:
Vy = V₀y + (-g) * t
where Vy is the vertical component of the velocity, V₀y is the initial vertical velocity component, g is the acceleration due to gravity, and t is the time.
0 = 1.60 × 10³ m/s * sin(43.0°) + (-9.81 m/s²) * t
Now, rearranging the equation, we find:
t = (1.60 × 10³ m/s * sin(43.0°)) / 9.81 m/s²
t ≈ 5.08 s
Since the shell reaches maximum height halfway through its flight time, the total flight time is twice the time it takes to reach maximum height:
Total flight time = 2 * 5.08 s
Total flight time ≈ 10.16 s
b) So, the shell is in motion for approximately 10.16 seconds. Just enough time for it to wreak havoc and have some mid-air fun!
To find the shell's horizontal range, we can first break down the initial velocity into its horizontal and vertical components. The horizontal component can be found by multiplying the initial velocity by the cosine of the angle:
Vx = V0 * cos(θ)
= 1.60 × 10^3 m/s * cos(43.0°)
Next, we can calculate the time it takes for the shell to reach its highest point in its trajectory. At this height, the vertical component of the velocity becomes zero. We can use the following formula to find this time:
Vf = Vi + at
Where:
Vf = final velocity in the vertical direction (which is 0 m/s)
Vi = initial velocity in the vertical direction (V0 * sin(θ))
a = acceleration due to gravity (-9.81 m/s^2)
t = time
0 = V0 * sin(θ) - 9.81 * t
Solving for t:
t = V0 * sin(θ) / 9.81
Once we have the time it takes to reach the highest point, we can double it to find the total time of flight. This is because the time taken to ascend is equal to the time taken to descend back to the ground.
Total time of flight = 2 * t
Finally, we can find the horizontal range by multiplying the horizontal velocity by the total time of flight:
Range = Vx * Total time of flight
Now we can calculate:
a) To find the shell's horizontal range:
Vx = 1.60 × 10^3 m/s * cos(43.0°)
≈ 1.16 × 10^3 m/s
t = (1.60 × 10^3 m/s) * sin(43.0°) / 9.81 m/s^2
≈ 5.09 s
Total time of flight = 2 * 5.09 s
≈ 10.2 s
Range = (1.16 × 10^3 m/s) * (10.2 s)
≈ 1.18 × 10^4 m
Therefore, the shell's horizontal range is approximately 1.18 × 10^4 meters.
b) To find the time of flight:
Total time of flight = 2 * t
≈ 2 * 5.09 s
≈ 10.2 s
Therefore, the shell is in motion for approximately 10.2 seconds.
To find the horizontal range of the shell, we need to analyze the horizontal and vertical components separately.
(a) Horizontal Range:
1. First, we need to find the shell's initial velocity components.
Given: Initial speed (v₀) = 1.60 × 10³ m/s
Initial angle (θ) = 43.0°
The horizontal component (v₀x) can be found using the formula: v₀x = v₀ * cos(θ)
v₀x = 1.60 × 10³ m/s * cos(43.0°)
2. Now, we can find the time of flight (T) of the shell using the vertical component.
The formula for the time of flight is given by: T = (2 * v₀y) / g
where v₀y is the vertical component of initial velocity.
The vertical component (v₀y) can be found using the formula: v₀y = v₀ * sin(θ)
v₀y = 1.60 × 10³ m/s * sin(43.0°)
3. Substituting the values into the formula, we get:
T = (2 * v₀y) / g
T = (2 * (1.60 × 10³ m/s * sin(43.0°))) / 9.81 m/s²
4. Once we have the value of T, we can find the horizontal range (R) using the formula: R = v₀x * T
The horizontal range is given by:
R = (1.60 × 10³ m/s * cos(43.0°)) * ((2 * (1.60 × 10³ m/s * sin(43.0°))) / 9.81 m/s²)
(b) Time of Flight:
Using the value of T obtained in step 3, we have the time of flight of the shell.
T = (2 * (1.60 × 10³ m/s * sin(43.0°))) / 9.81 m/s²
Now, we can substitute the values into the equations to find the answers.
(a) R = (1.60 × 10³ m/s * cos(43.0°)) * ((2 * (1.60 × 10³ m/s * sin(43.0°))) / 9.81 m/s²)
Calculate this expression to find the horizontal range.
(b) T = (2 * (1.60 × 10³ m/s * sin(43.0°))) / 9.81 m/s²
Calculate this expression to find the time of flight.