An object projected at an angle to a ground level has a time of flight 4 seconds to move through still air. Calculate the maximum height attained by the object.[g = 10ms\(^{-2}\)]
Given: time of flight T = 4s, g = 10ms\(^{-2}\), H\(_{max}\)
\(\theta\) = ?
T = \(\frac{2u sin\theta}{g}\)
4 = \(\frac{2 \times u sin \theta}{10}\)
40 = 2u sin\(\theta\)
sin\(\theta\) = \(\frac{20}{u}\)
But, H\(_{max}\) = \(\frac{u^2 sin^2 \theta}{2g}\)
H\(_{max}\) = \(\frac{u^2 \times (\frac{20}{u})^2}{2 \times 10}\)
H\(_{max}\) = \(\frac{20 \times 20}{20}\) = 20 metres.(u\(^2\)s cancel out)
Thus H\(_{max}\) attained = 20metres.
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