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WAEC SSCE - Physics - 2013 (Objective)

Question 1 Report

A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms\(^{-2}\)]. Calculate the speed of projection

Answer Details

To determine the speed of projection of a body projected vertically upwards in a vacuum, we can use the principles of kinematics and the given information. Here are the steps:

  1. Understand the total time of flight: The total time taken for the body to go up and come back down is 1.2 seconds.

  2. Determine the time to reach the maximum height: Since the time to go up is equal to the time to come down, the time to reach the maximum height is half of the total time of flight.

    t up = 1.2  s 2 = 0.6  s t_{\text{up}} = \frac{1.2 \text{ s}}{2} = 0.6 \text{ s}
  3. Use the kinematic equation for velocity: The velocity at the maximum height is 0 (since the body momentarily stops before falling back down). We can use the equation:

    v = u − g t v = u - gt

    where v v is the final velocity (0 m/s at the maximum height), u u is the initial velocity (speed of projection), g g is the acceleration due to gravity (10 m/s²), and t t is the time taken to reach the maximum height (0.6 s).

    Setting v = 0 v = 0 :

    0 = u − g ⋅ t 0 = u - g \cdot t
  4. Solve for u u :

    u = g ⋅ t = 10  m/s 2 ⋅ 0.6  s = 6  m/s u = g \cdot t = 10 \text{ m/s}^2 \cdot 0.6 \text{ s} = 6 \text{ m/s}

Therefore, the speed of projection is 6.0  m/s \boxed{6.0 \text{ m/s}} 6.0 m/s