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.
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for θ max hight = v²sin²θ/2g = h₁
for 90-θ = v²cos²θ/2g = h₂
| h1-h2 | =50
v²/2g(cos²© - sin²©) = 50
for 90-θ = v²cos²θ/2g = h₂
| h1-h2 | =50
v²/2g(cos²© - sin²©) = 50
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