Question 14
- A particle is projected vertically upwards from a point O with a velocity of 75ms-1. Find the:
(i) velocity of the particle at the end of 5 seconds;
(ii) height attained when the velocity is 15 ms-1 ;
(iii) the times when the particle is 270 m above O.
⦋Take g = 10ms-2⦌
(b) A force of 24 N acts on a mass q kg and increases its speed from 6 ms-1 to 42 ms-1 in 6 seconds. Find the value of q.
Observation
The Chief Examiner reported that most candidates performed very well in this question.
In part (ai) as required, they used v = u – gt and substituting u = 75, g = 10 and t = 5 to have v = 75 – 10(5) = 25 ms-1. In part(aii), using v2 = u2 + 2as, where v = 15 ms-1, we have
S = 
270 m and in part(aiii), 270 = 75t - 
10t2 and simplifying to get t2 – 15t + 54 = 0 and factorizing to solve for t gives t = 6 seconds or t = 9 seconds.
In Part (b), the majority of the candidates handled the problem so well. This they did by first finding acceleration (a) = 
.
Thereafter, using q = 
.