A gun is mounted at the top of light house of height h. A similar gun is mounted in the boat in the sea. Both the guns can fire shots with speed upto under root 2gk, where k>h. What will be the area in which gun at the top can be gun in the boat, but gun in the boat cannot be gun at the top?
Well
To solve this problem, we can break it down into two parts:
Part 1: Determine the range of the gun mounted on the top of the lighthouse.
Part 2: Determine the range of the gun mounted on the boat.
Part 1:
The range of a projectile launched at an angle θ with an initial speed v can be calculated using the formula:
R = (v^2 * sin(2θ))/g
In this case, the gun on top of the lighthouse will have an initial speed up to √(2gk), where k > h. Assuming the gun is fired horizontally (θ = 0), the equation becomes:
R_top = (√(2gk)^2 * sin(2*0))/g
= (2gk * sin(0))/g
= 0
Since the range R_top is zero, it means that the gun on top of the lighthouse cannot hit any target horizontally.
Part 2:
Similarly, let's calculate the range of the gun mounted on the boat. Given that the boat is at sea level, the height of the boat gun is h.
To achieve the maximum range, the gun should be fired at an angle of 45 degrees (θ = 45°).
R_boat = (√(2gk)^2 * sin(2*45))/g
= (√(2gk)^2 * sin(90))/g
= (√(2gk)^2 * 1)/g
= 2gk/g
= 2k
The range of the gun mounted on the boat is 2k.
To find the area in which the boat gun can reach, but the top gun cannot, we subtract the area covered by the lighthouse gun from the area covered by the boat gun.
Area = π * (2k)^2 - π * (0)^2
= 4πk^2
Therefore, the area in which the gun on top of the lighthouse cannot reach, but the gun on the boat can, is 4πk^2.