A shell is fired from the ground with an initial speed of 1.58 103 m/s at an initial angle of 31° to the horizontal.
(a) Neglecting air resistance, find the shell's horizontal range.
m
(b) Find the amount of time the shell is in motion.
s
(a) Well, before we dive into the math, let's take a moment to appreciate that this shell is fired without any concern for air resistance. Talk about living life on the edge! Now, let's calculate the horizontal range.
To find the range, we need to determine how long the shell stays in the air. We can do this by finding the time it takes for the shell to hit the ground. Just don't ask the shell how it feels about "hitting rock bottom."
The horizontal component of the initial velocity is given by v₀x = v₀ * cos(θ), where v₀ is the initial speed and θ is the launch angle. Plugging in the values, we have v₀x = 1.58 * 10³ * cos(31°).
Next, we need to find the time it takes for the shell to hit the ground. The equation for displacement in the horizontal direction is x = v₀x * t, where x is the range and t is time. We know that when the shell hits the ground, its vertical displacement is equal to zero.
Since gravity acts in the vertical direction, we can use the equation Δy = v₀y * t + (1/2) * g * t², where v₀y is the initial vertical velocity, g is the acceleration due to gravity, and Δy is the change in height.
In this case, the initial vertical velocity is given by v₀y = v₀ * sin(θ). Since the shell starts and ends at the same height (no change in height), Δy is equal to zero.
Therefore, we can rewrite the equation as 0 = v₀y * t + (1/2) * g * t². Plugging in the values, we have 0 = 1.58 * 10³ * sin(31°) * t + (1/2) * 9.8 * t². Rearranging the equation gives us t² + (2 * (1.58 * 10³ * sin(31°)) / 9.8) * t = 0. You can solve this quadratic equation to find the time t when the shell hits the ground.
Once you have the time, you can multiply it by the horizontal component of the initial velocity to find the range: Range = v₀x * t. Crunch the numbers, and you'll have your horizontally impressive range (in meters)!
(b) Now that we've calculated the range, let's move on to how long this adventurous shell stays in the air. We already know its horizontal range, but the time is ticking, so let's get the timing right!
To find the total time of flight, we can use the equation t = 2 * t_max, where t_max is the time it takes for the shell to reach its maximum height.
The time to reach the maximum height can be found using the equation Δy = v_y * t + (1/2) * g * t². Here, Δy = 0 (since at the maximum height, the vertical displacement is zero), and v_y = v₀y. With some algebraic maneuvers, you can solve for t_max.
Once you have t_max, you can double it to find the total time of flight. Now that we've taken the time (and double-checked our calculations), we have the answer to how long this shell is in motion (in seconds)!
Remember, though, despite all the math, this shell still knows how to make an explosive entrance!
To find the shell's horizontal range and time of flight, we can use the equations of projectile motion.
(a) To find the range, we can use the equation:
Range = (initial velocity *)^2 * sin(2θ) / g
where:
- initial velocity * is the initial speed of the shell
- θ is the initial angle to the horizontal
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the given values:
Range = (1.58 * 10^3 m/s)^2 * sin(2 * 31°) / 9.8 m/s^2
Calculating this:
Range = (2.4964 * 10^6 m^2/s^2) * 0.5236 / 9.8 m/s^2
Range ≈ 1.342 * 10^5 m
Therefore, the shell's horizontal range is approximately 1.342 * 10^5 meters.
(b) To find the time of flight, we can use the equation:
Time of flight = 2 * (initial velocity * * sin(θ)) / g
Plugging in the given values:
Time of flight = 2 * (1.58 * 10^3 m/s * sin(31°)) / 9.8 m/s^2
Calculating this:
Time of flight = (2 * 816.8 m/s) / 9.8 m/s^2
Time of flight ≈ 165.92 s
Therefore, the amount of time the shell is in motion is approximately 165.92 seconds.
To solve this problem, we can use the basic equations of projectile motion. Let's break down the problem step by step.
(a) To find the shell's horizontal range, we need to determine the distance traveled by the shell in the horizontal direction before hitting the ground.
The range of a projectile can be calculated using the formula:
Range = (initial velocity^2 * sin(2θ)) / g
where:
- initial velocity is the initial speed of the shell (1.58 * 10^3 m/s)
- θ is the launch angle (31 degrees)
- g is the acceleration due to gravity (9.8 m/s^2)
Plugging in the values into the formula, we can find the range:
Range = (1.58 * 10^3 m/s)^2 * sin(2 * 31°) / 9.8 m/s^2 = (2.5004 * 10^6 m^2/s^2) * sin(62°) / 9.8 m/s^2
Using a calculator, we can calculate the value inside the parenthesis and then multiply it by the sine of 62° and divide by 9.8 to get the range. The result will be in meters.
(b) To find the time the shell is in motion, we can use the equation:
Time of flight = (2 * initial velocity * sin(θ)) / g
Again, plugging in the given values, we have:
Time of flight = (2 * 1.58 * 10^3 m/s * sin(31°)) / 9.8 m/s^2
Using a calculator, we can calculate the sine of 31°, then multiply it by 2 and the initial velocity, and finally divide it by gravity (9.8 m/s^2). The result will be in seconds.
By following these steps, you can find the horizontal range and time of flight for the shell.