Physics (Essay) Paper 2 Nov/Dec 2015

Question 2

 

Part 1:  Candidates were required to answer any five questions from this part.

           

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 s, calculate the maximum attainable height.
( g = 10 ms-2 )

Observation

 

Most candidates identified the demand of this question.  Most of the candidates answered it using the third equation of motion and was well answered.  Few that used the projectile equation approach had difficulty because of their inability to manipulate sin2 30.
Some candidates used the formula for calculating range instead of that for calculating maximum height.

The expected answer is :

H  =  (u sin)t – ½ gt2              OR    H  =   U2 sin2 2g                            
=  (42 sin30)2.1 – ½ x 10 x (2.1)2            =   422 x (sin 30)2

         =22.05m                            2 x 10=

                                           22.05 m