A body is projected with velocity $30 \mathrm{~m} / \mathrm{s}$. The time taken by the body to reach

Question

A body is projected with velocity $30 \mathrm{~m} / \mathrm{s}$. The time taken by the body to reach at highest point is (angle of projection is $30^{\circ}$ )

Professor PC Thomas Classes · Answer

Time of ascent $t = \frac{u\sin	heta}{g} = \frac{30\sin 30^{\circ}}{10} = \frac{30(0.5)}{10} = 1.5 \mathrm{s}$.