Question

A projectile fired at \(30^{\circ}\) to the ground is observed to be at the same height at time 3s and 5s after projection, during its flight. The speed of projection of the projectile is_____ ms\(^{-1}\).

Hint:

For projectile motion, the total time of flight can be found by adding the times at which the projectile is at the same height, and using the kinematic equations to solve for the initial velocity.

Answer 1

Correct Answer: 80

The time of flight for a projectile is given by the formula: \[ T = \frac{2u \sin \theta}{g} \] where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. From the given problem, the projectile returns to the same height at 3s and 5s. Thus, the total time of flight is: \[ T = 5 + 3 = 8 \, \text{seconds} \] Substituting the values into the formula for time of flight: \[ 8 = \frac{2u \sin 30^{\circ}}{10} \] \[ 8 = \frac{2u \times \frac{1}{2}}{10} \] \[ 8 = \frac{u}{10} \] \[ u = 80 \, \text{m/s} \] Thus, the speed of projection is 80 m/s.