A lead shot is projected from tha ground level with a velocity u at an angle \(\theta\) to the horizontal. Given the time, t for the lead shot to reach its maximum height as; t = \(\frac{u \sin \theta}{g}\) where "g" is the acceleration of free fall due to gravity, show that the greatest height reached by the body is h\(_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}\)
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Correct Answer: Option n
Explanation:
For verticsl motion of projectile h = (U sin \(\theta\)) t \(\frac{1}{2}\)gt\(^2\)
= U sin \(\theta\) (\(\frac{Usin \theta}{g}\))\(\frac{1}{2}\)g (\(\frac{Usin \theta}{g}\)) = (\(\frac{U^2 \sin^2 \theta}{g}\))
For verticsl motion of projectile h = (U sin \(\theta\)) t \(\frac{1}{2}\)gt\(^2\)
= U sin \(\theta\) (\(\frac{Usin \theta}{g}\))\(\frac{1}{2}\)g (\(\frac{Usin \theta}{g}\)) = (\(\frac{U^2 \sin^2 \theta}{g}\))