math

For a projectile launched from the ground at an angle of 45°, with initial velocity V₀, the time of flight, t, is given by: t = (2V₀sin45°)/g The range, R, of the projectile is as follows: R = (V₀²sin2θ)/g In this case, θ is 45°. We also know that the height of the tower is 550 feet, and the distance between the tower and the initial firing location is 600 ft. Therefore, we are looking for the initial velocity, V₀, that is required to cover a distance of 600 ft and a height of 550 ft. From the two equations above, we can solve for V₀. First, we solve for the time of flight t: t = (2V₀sin45°)/g → V₀ = (gt)/2sin45° Next, we can substitute this into the equation for range: R = (V₀²sin2θ)/g → (gt/2sin45°)² sin2(45°) = 600 g = (4sin²45°)R/t² → g = (4sin²45°)(600)/t² Finally, we can substitute this