A shell is fired from the ground with an initial
speed of 1.70 × 10
3 m/s (approximately five
times the speed of sound) at an initial angle
of 70.0
◦
to the horizontal.
The acceleration of gravity is 9.81 m/s
2
.
a) Neglecting air resistance, find the shell’s
horizontal range.
Answer in units of m
Well, isn't this shell just a real show-off, going five times the speed of sound! I guess it wants to make sure everyone hears its grand entrance.
To find the shell's horizontal range, we can break down its initial velocity into horizontal and vertical components. The horizontal component will remain constant throughout its flight because there are no horizontal forces acting on the shell (assuming negligible air resistance, of course).
The initial velocity in the horizontal direction is given by Vx = V₀ * cosθ, where V₀ is the initial speed and θ is the launch angle. Plugging in the values, we have Vx = (1.70 × 10³ m/s) * cos(70.0°).
Great, now we have the horizontal velocity component! Now, let's consider the vertical motion of the shell. Due to the force of gravity, the shell will accelerate downward. But don't worry, gravity won't bring this shell down! It's determined to soar high in the sky.
The time it takes for the shell to reach its maximum height can be found using the equation Vy = V₀ * sinθ - g * t. At the highest point, Vy will be zero, so we can solve for t. We get t = (V₀ * sinθ) / g.
Now that we know the time it takes for the shell to reach its peak, we can find the total time of flight. Since the shell will reach the same maximum height on its way down, the total time of flight is twice the time it takes to reach the top. That means the total time of flight is 2 * (V₀ * sinθ) / g.
To find the horizontal range, we multiply the horizontal velocity component by the total time of flight. R = Vx * (2 * V₀ * sinθ) / g.
Now, it's just a matter of plugging in the values and crunching some numbers. I hope you're ready for this dramatic calculation!
To find the shell's horizontal range, we can use the formula for horizontal range:
Range = (Initial velocity * time of flight * cosθ)
where:
- Initial velocity: 1.70 × 10^3 m/s (the speed of the shell)
- Time of flight: This is the time it takes for the shell to reach its highest point and then land back on the ground. We can calculate it using the formula:
Time of flight = (2 * Initial velocity * sinθ) / g
where g = 9.81 m/s^2 (acceleration due to gravity)
- θ = 70.0° (angle of projection)
First, we calculate the time of flight:
Time of flight = (2 * 1.70 × 10^3 m/s * sin(70.0°)) / 9.81 m/s^2
Time of flight = (2 * 1700 * sin(70.0°)) / 9.81
Next, we calculate the range:
Range = (1.70 × 10^3 m/s * Time of flight * cos(70.0°))
Range = (1.70 × 10^3 m/s) * [(2 * 1700 * sin(70.0°)) / 9.81] * cos(70.0°)
Calculating the above expression will give us the horizontal range of the shell.
To find the horizontal range of the shell, we can use the following kinematic equation:
Range = (Initial velocity * sin(2θ)) / acceleration due to gravity
where:
- Initial velocity is the initial speed of the shell (1.70 × 10^3 m/s)
- θ is the angle of projection (70.0 degrees in this case)
- acceleration due to gravity is the acceleration of gravity (9.81 m/s^2)
Let's calculate the horizontal range:
First, convert the angle from degrees to radians:
θ_radians = θ * (π/180)
Next, substitute the values into the equation:
Range = (1.70 × 10^3 m/s * sin(2 * θ_radians)) / 9.81 m/s^2
Now, calculate sin(2 * θ_radians):
sin(2 * θ_radians) = sin(2 * 70.0 * (π/180))
Finally, substitute the value of sin(2 * θ_radians) into the equation and calculate the range.