Question

A projectile fired at 30° to the ground is observed to be at same height at time 3 s and 5 s after projection, during its flight. The speed of projection of the projectile is m s−1 . (Given g=10 m s−2 )

Answer 1

80

Answer 2

The times $t_1 = 3$ s and $t_2 = 5$ s, when the projectile is at the same height, are symmetric around the time of maximum height ($t_{peak}$). $t_{peak} = \frac{t_1 + t_2}{2} = \frac{3 \text{ s} + 5 \text{ s}}{2} = 4 \text{ s}$ The time to reach maximum height is also given by the formula: $t_{peak} = \frac{u \sin \theta}{g}$ Given the projection angle $\theta = 30^\circ$ and $g = 10$ m/s$^2$: $4 \text{ s} = \frac{u \sin 30^\circ}{10 \text{ m/s}^2}$ Since $\sin 30^\circ = \frac{1}{2}$: $4 = \frac{u \times \frac{1}{2}}{10}$ $4 = \frac{u}{20}$ Solving for the speed of projection $u$: $u = 4 \times 20$ $u = 80 \text{ m/s}$