An artillery gun is trying to hit a target that is 7.2 km away. If the shell is fired at 680 ms-1 at what angle above the horizontal should the gun be fired to hit the target assuming that air resistance is negligible? [g=9.8 ms-2]
To solve this problem, we can use the equations of projectile motion. The projectile will follow a curved trajectory, and we need to find the angle at which it should be fired to hit the target 7.2 km away.
Let's break down the problem step by step:
1. Define the given values:
- The initial velocity of the shell (u) = 680 ms⁻¹
- The horizontal distance to the target (range) = 7.2 km = 7200 m
- The acceleration due to gravity (g) = 9.8 ms⁻²
2. Convert the range to the vertical component:
The range of the projectile is the horizontal distance traveled, which is affected by the initial velocity and the angle of projection. We need to consider the vertical component of the range.
Let's assume the angle of projection is θ. Then the vertical component of the range is given by:
Range_y = Range * sin(θ)
Range_y = 7200 m * sin(θ)
3. Determine the time of flight:
The time taken by the projectile to reach its target can be found using the equation:
range = (u^2 * sin(2θ)) / g
Substituting the given values:
7200 = (680^2 * sin(2θ)) / 9.8
4. Solve for the angle (2θ):
Rearranging the equation, we get:
sin(2θ) = (7200 * 9.8) / (680^2)
sin(2θ) ≈ 0.2366
We can use inverse sine to find 2θ:
2θ ≈ arcsin(0.2366)
2θ ≈ 13.4 degrees
5. Finally, find the angle (θ):
Divide 2θ by 2 to get the angle above the horizontal:
θ = 13.4 degrees / 2
θ ≈ 6.7 degrees
So, the artillery gun should be fired at an angle of approximately 6.7 degrees above the horizontal to hit the target 7.2 km away, assuming air resistance is negligible.