An artillery shell is launched on a flat, horizontal field at an angle of α = 36.2° with respect to the horizontal and with an initial speed of v0 = 318 m/s. What is the horizontal distance covered by the shell after 7.68 s of flight?
I'm confused on how to even begin this problem.
horizontal speed is
318 cos 36.2° = 256.613 m/s
So, how far does it travel in 7.68s?
Remember that the horizontal speed does not change after takeoff. The only force acting is the vertical pull of gravity.
To solve this problem, you can use the kinematic equations of motion to determine the horizontal distance covered by the artillery shell. Let's break down the problem step by step:
1. Start by splitting the initial velocity of the shell into its horizontal (v0x) and vertical (v0y) components. The horizontal component, v0x, is given by v0 * cos(α), where α is the launch angle (36.2°) and v0 is the initial velocity (318 m/s). So, v0x = 318 m/s * cos(36.2°).
2. Next, determine the vertical component, v0y, of the initial velocity. This can be calculated as v0 * sin(α), where sin(α) is the sine of the launch angle. So, v0y = 318 m/s * sin(36.2°).
3. Since the shell is launched on a flat, horizontal field, there is no vertical acceleration (assuming negligible air resistance). Therefore, the time of flight (t) for the shell will be the same in the horizontal and vertical directions. In this case, the time of flight is given as 7.68 seconds.
4. To find the horizontal distance covered by the shell, you can use the equation x = v0x * t, where x is the horizontal distance and t is the time of flight. Substitute the values you have: x = (318 m/s * cos(36.2°)) * 7.68 s.
5. Calculate the horizontal distance x using a calculator or by hand.
By following these steps, you should be able to determine the horizontal distance covered by the artillery shell after 7.68 seconds of flight.