A cricket ball is thrown at a speed of $28 ext{ m/s}$ in a direction $30^\circ$ above the horizonta

Question

A cricket ball is thrown at a speed of $28	ext{ m/s}$ in a direction $30^\circ$ above the horizontal. What is the maximum height attained by the ball?

Professor PC Thomas Classes · Answer

Maximum height $h_{max} = \frac{u^2\sin^2	heta}{2g} = \frac{(28\sin30^\circ)^2}{2(9.8)} = \frac{14^2}{19.6} = \frac{196}{19.6} = 10.0	ext{ m}$.