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Physics Paper 2, Nov/Dec. 2009  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments
Weakness/Remedies
Strength
















Question 2
Part I:      Candidates were required to answer any five questions from this part.
  1. A projectile is launched with a velocity of 42 ms-1 at an angle of 30o to the horizontal.  If the time of flight is 4.2 seconds, calculate the maximum attainable height.  [g = 10 ms-2]

       

     

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Observation

This question was a popular one among the candidates and was correctly solved by many candidates.  However few candidates exhibited poor computational skills in the handling of square and square roots of angles in the formula used.  Some even used wrong formula.

The expected answer is:

The projectile takes 2.1 seconds to reach the maximum height H.                              
H  =  (U sinθ) t  - ½ qt2                                                                                                            

=   (42 sin  30) (2.1) - ½  x 10  x (2.1)2                                                                                        
=    44.1   -   22.05
=    22.05m                                                                                                                    

            OR
               
H  =  U2 sin2θ
              2g                                                                                                                    
 =   422  x ( sin  30)2
 2 x 10                                                                                                                

=   22.05m 
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