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Physics Paper 2, Nov/Dec. 2013  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments
Weakness/Remedies
Strength
















Question 1

          A particle is projected as an angle of 30o to the horizontal with a speed of 250 ms-1
          Calculate  the:   (a)   time of flight of the particle;
                                    (b)   speed of the particle at its maximum height.
                                            [ g  =  10ms-2]
          Part (a). This was a very popular question among the candidates and it was fairly well attempted.  However, some candidates started off with wrong formula while some settled with time t =   instead of the correct formula  T =
 Part (b).  This question was poorly attempted.  Candidates should note that the vertical component of the velocity at the maximum height is zero, since it stops moving upwards.  However, horizontal component of the velocity remains the same i.e V(x) = 250 cos 30 = 216.5ms-1

 

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comments

The expected answers are:
 (a).       T          =                                                                                                  
                          =                                                                                        
                          =          25 s                                                                                        
 
 (b)        At maximum height, speed, v = vx = 250 cos 30o          

                                                                       = 216.5 ms-1  
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