Derive an expression for the maximum height of a projectile launched at an angle

Question

Derive an expression for the maximum height of a projectile launched at an angle of theta = 45 degree in terms of the range R (the distance traveled on flat ground). From the kinematics equations, the maximum height of a projectile launched at theta = 45 degree is H max = The range of a projectile, launched at any angle theta, is R = The time that passes before a projectile, launched at any angle theta, returns to its original height is The time that passes before a projectile, launched at any angle theta, returns to its original height is Delta t = When theta =45 degree, R = So, when theta = 45 degree, H max =

Explanation / Answer

part 1: (vf)^2 = (vi)^2 + 2g(H) at max height, the final velocity would be zero 0 = (vi*sin45)^2 + 2gH 0 = 0.5vi^2 + 2gH 2gH = 0.5vi^2 4gH = vi^2 H = (vi)^2 / 4g (answer to part 1 is 4g or 4*9.8 = 39.2) part 2: R = (vi*cos(theta))*(delta t) so the answer for part 2 is (cos) and (t) part 3: delta y = (vi*sin(theta))(t) + 0.5(g)(t^2) 0 = (vi * sin(theta)) + 0.5gt 0.5gt = (vi * sin(theta)) t = (2)(vi * sin (theta)) / g so the answers for part 3 should be: 2 sin(theta) g OR 9.8 Part 4: R = (vi * cos(theta)) * (2 vi sin(theta)) / g R = (vi)^2/g so the answer to part 4 is g or 9.8 part 5: H = (1/4)(R) so the answer to part 5 should be: (1/4)(R) BOL

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