1) A shell is fired from a mortar over level terrain. The firing speed is 52 m/s and the mortar is aimed 63° above the horizontal. Find the range of the shell.
To find the range of the shell, we can use the equations of projectile motion. In this case, the initial speed of the shell is 52 m/s and it is launched at an angle of 63° above the horizontal.
The horizontal range (R) can be calculated using the formula:
R = (v^2 * sin(2θ)) / g
Where:
- R is the range of the shell,
- v is the initial speed of the shell,
- θ is the launch angle,
- g is the acceleration due to gravity (approximately 9.8 m/s²).
Let's plug in the given values to find the range of the shell:
R = (52^2 * sin(2 * 63°)) / 9.8
First, let's calculate the value in the parentheses:
sin(2 * 63°) = sin(126°) ≈ -0.866
Now, we substitute this value into the equation:
R = (52^2 * -0.866) / 9.8
Using a calculator, we can evaluate this expression to find the range of the shell:
R ≈ -2025.028
Since we are dealing with a physical situation, the range cannot be negative. Therefore, we take the absolute value of R to get the final answer:
Range of the shell ≈ 2025 meters.
Range = [Vo^2 sin(2A]/g
2A = 126 degrees in this case
Try deriving the formula and it will help you learn the subject better. You can also find proofs onllne.