1. A particle is projected vertically upwards with a speed of 25 ms-1 from a point on the ground. Find the:
2. position of the particle after 4 seconds;
3. maximum height reached;
4. time taken to reach the maximum height;
5. time when the particle is 30m above the ground.

[Take g = 10 ms-2].

1. Calculate the force which acts on a body of mass 3 kg moving at 2.5 ms-1 for 0.5 seconds, if the final velocity is 4.5 ms-1.

This question demanded the application of Newton’s equations of motion. Majority of the candidates were reported not to attempt this question and the performance of the few who attempted it was described as fair.
In part (a), candidates were expected to apply the formula s = ut + at2. Here, u = 25 ms-1, t = 4 seconds and a = -g = -10 ms-2. Therefore, s = 25(4) – 5(16) = 20 m. At maximum height, v = 0. Hence using v2 = u2 + 2as, 0 = (25)2 – 20(s). This implied that s =  = 31.25 m. The time taken to reach the maximum height was gotten by using the formula v = u - gt where v = 0, u = 25 ms-1 and g = 10 ms-2. i.e. 0 =25 -10t. Thus t =  = 2.5 seconds. The time when the particle was 30 m above the ground was gotten using the formula s = ut + at2. Here, 30 = 25t - (10)t2. i.e. 30 = 25t – 5t2 or t2 – 5t + 6 = 0. Solving this quadratic equation gave t = 2 seconds and 3 seconds.
In part (b), candidates were expected to apply the formula Ft = m(v – u). Here, v = 4.5 ms-1, u = 2.5 ms-1, m= 3 kg and t = 0.5seconds. Therefore, F =  = 12 N.

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