Question 1 Report
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\(_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}\)