A trench mortar fires a shell at an angle of 30° with the horizontal and with a speed of 400 m/s. Find its range and the maximum height it attains. (a)Range (b) Height
if you don't have the formulas handy, you can find them and their derivations in the wikipedia article on Trajectory.
vertical speed = 200 = Vi
m g h = (1/2) m Vi^2
solve for h = 200^2 / (2g) = max height
find time to top
v = Vi - g t
0 at top so
t = 200 / g at top
time in air = 2 t = 400/g
horizontal speed = 400 cos30 = u
range = u * 400/g
To find the range and maximum height of a projectile fired at an angle, we can use the equations of projectile motion. The two main equations we'll use are:
1. Range = (Initial velocity^2 * sin(2θ)) / g
2. Maximum height = (Initial velocity^2 * sin^2(θ)) / (2*g)
Where:
- Initial velocity is the speed at which the shell is fired (given as 400 m/s).
- θ (theta) is the launch angle (given as 30°).
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now let's plug in the values and calculate the range and maximum height:
(a) Range:
Range = (400^2 * sin(2*30°)) / 9.8
To find sin(60°) (which is sin(2*30°)), we can use the identity sin(2θ) = 2*sin(θ)*cos(θ).
sin(60°) = 2 * sin(30°) * cos(30°)
sin(60°) = 2 * (1/2) * √3/2
sin(60°) = √3/2
Range = (400^2 * √3/2) / 9.8
Now calculate the value using a calculator:
Range = (160,000 * √3) / 9.8
Range ≈ 9130.92 meters
Therefore, the range of the shell is approximately 9130.92 meters.
(b) Maximum Height:
Maximum height = (400^2 * sin^2(30°)) / (2 * 9.8)
To find sin^2(30°), we can use the identity sin^2(θ) = 1 - cos^2(θ).
sin^2(30°) = 1 - cos^2(30°)
sin^2(30°) = 1 - (√3/2)^2
sin^2(30°) = 1 - 3/4
sin^2(30°) = 1/4
Maximum height = (400^2 * 1/4) / (2 * 9.8)
Maximum height = (160,000 * 1/4) / (2 * 9.8)
Maximum height = (40,000) / (19.6)
Maximum height ≈ 2040.82 meters
Therefore, the maximum height the shell attains is approximately 2040.82 meters.
In summary:
(a) Range ≈ 9130.92 meters
(b) Maximum height ≈ 2040.82 meters