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})
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}))