An artillery shell is fired at an angle of 32.2◦ above the horizontal ground with an initial speed of 1610 m/s.
The acceleration of gravity is 9.8 m/s
2.
Find the total time of flight of the shell,
neglecting air resistance.
Answer in units of min
Find its horizontal range, neglecting air resistance.
Answer in units of km
You can do this yourself,
The time of flight is twice the time it takes for the vertical velocity component to be come zero.
The horizontal range is
2 sin32.2 cos 32.2*(Vo^2/g)
= (Vo^2/g) sin 64.4 = ____
then how would you find vertical velocity component?
To find the total time of flight of the shell, we need to consider the vertical motion of the shell. We can use the kinematic equation for vertical motion:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
- y is the vertical displacement of the shell (in this case, it will be 0 because the shell starts and ends at ground level)
- y0 is the initial vertical position of the shell (which is also 0)
- v0y is the vertical component of the initial velocity of the shell
- g is the acceleration due to gravity (which is a negative value because it acts downward)
- t is the time
We know that the angle of projection is 32.2° above the horizontal, and the initial speed of the shell is 1610 m/s. We can calculate the vertical component of the initial velocity as follows:
v0y = v0 * sin(θ)
where θ is the angle of projection.
v0y = 1610 m/s * sin(32.2°)
Now, we can substitute the values into the kinematic equation:
0 = 0 + (1610 m/s * sin(32.2°)) * t - (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation:
0 = 536.008 m/s * t - 4.9 m/s^2 * t^2
Rearranging the equation:
4.9 m/s^2 * t^2 = 536.008 m/s * t
Dividing both sides by t:
4.9 m/s^2 * t = 536.008 m/s
Simplifying:
t = 536.008 m/s / 4.9 m/s^2
t = 109.39 s
To convert the time to minutes, we divide by 60:
Total time of flight = 109.39 s / 60 = 1.8232 min (rounded to 4 decimal places)
Now let's calculate the horizontal range of the shell. The horizontal range is the horizontal displacement of the shell, which is given by:
range = v0x * t
where v0x is the horizontal component of the initial velocity of the shell.
We can calculate the horizontal component of the initial velocity as follows:
v0x = v0 * cos(θ)
where θ is the angle of projection.
v0x = 1610 m/s * cos(32.2°)
Now, substituting the values into the equation:
range = (1610 m/s * cos(32.2°)) * 109.39 s
Now, convert the range from meters to kilometers by dividing by 1000:
Horizontal range = (1610 m/s * cos(32.2°)) * 109.39 s / 1000
Horizontal range = 57.47 km (rounded to 2 decimal places)
Therefore, neglecting air resistance, the total time of flight of the shell is approximately 1.8232 min and its horizontal range is approximately 57.47 km.