A projectile is launched with an initial speed u such that its horizontal range is exactly equal to its maximum height. What is the speed of the projectile at the highest point of its trajectory?

Step-by-step solution

The horizontal range R and maximum height H of a projectile are given by: R = u^2 2 g = 2u^2 g H = u^2 ^2 2g Given that R = H : 2u^2 g = u^2 ^2 2g Simplifying this gives: 2 = 2 = 4 From = 4 , we can find using a right-angled triangle with opposite side 4 and adjacent side 1 . The hypotenuse is 4^2 + 1^2 = 17 . Thus, = 1 17 . The speed at the highest point is the horizontal component of the initial velocity, as the vertical component becomes zero: v = u = u 17 Answer: u 17