An artillery shell is launched on a flat, horizontal field at an angle of a = 43.5° with respect to the horizontal and with an initial speed of v0 = 263 m/s. What is the horizontal distance covered by the shell after 6.77 s of flight?
What is the height of the shell at this moment?
To find the horizontal distance covered by the shell and the height at a specific time, we can use the kinematic equations.
First, let's find the horizontal distance covered by the shell after 6.77 s of flight.
The horizontal distance (x) covered by the shell can be calculated using the equation:
x = v0 * t * cos(a)
Given:
v0 = 263 m/s (initial speed of the shell)
t = 6.77 s (time of flight)
a = 43.5° (angle of launch)
First, we need to convert the angle from degrees to radians, as trigonometric functions in most programming languages expect inputs in radians. So, let's convert the angle:
a_rad = a * (π / 180)
= 43.5° * (π / 180)
≈ 0.759 rads
Now, we can plug in the values into the equation to find the horizontal distance:
x = 263 m/s * 6.77 s * cos(0.759 rads)
Using a calculator, we can compute:
x ≈ 1234.85 m
So, the horizontal distance covered by the shell after 6.77 s of flight is approximately 1234.85 meters.
Now, let's find the height of the shell at this moment.
The equation for height (y) in projectile motion is given by:
y = v0 * t * sin(a) - (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Let's plug in the values:
y = 263 m/s * 6.77 s * sin(0.759 rads) - (1/2) * 9.8 m/s^2 * (6.77 s)^2
Calculating this using a calculator:
y ≈ 744.24 m
Therefore, the height of the shell at this moment is approximately 744.24 meters.
t = 6.77 s
X = Vx*t , where
Vx = Vo*cos43.5 = 190.8 m/s
Y = Vyo*t - (g/2) t^2, where
Vyo = Vo*sin43.5 = 181.04 m/s and
g = 9.81 m/s^2
X and Y are your two answers. Proceed with the calculation.