Question
A person throws a ball from ground level at speed u and
angle θ. The ball reaches maximum height H and range R. What is the ratio H/R?Solution
Given:
- Maximum height: H = (u²·sin²θ) / (2g)
- Range: R = (u²·sin2θ) / g
= sin²θ / (2·sin2θ) Now use the identity: sin2θ = 2·sinθ·cosθ So, H/R = sin²θ / [2·(2·sinθ·cosθ)] = sinθ / (4·cosθ) = (1/4)·tanθ
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