Question

A ball is thrown at an angle theta and another ball is thrown at an angle (90-theta) with the horizontal direction from the same point wih velocity 98m/s. The second ball reaches 50m higher than the first ball. Find their individual heights.

Answer 1

for θ max hight = v²sin²θ/2g = h₁
for 90-θ = v²cos²θ/2g = h₂

| h1-h2 | =50
v²/2g(cos²© - sin²©) = 50