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 Physics Paper 2, Nov/Dec. 2011
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
Weakness/Remedies
Strength

Question 1

1. A stone is projected vertically upward with a speed of 30ms-1 from the top of a tower
of height 50m.  Neglecting air resistance, determine the time of flight on reaching the
ground.[g  =  10ms-2]

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Observation

This question on projectile motion was very popular among the candidates and was fairly well attempted.  However, some candidates did not understand the difference between  projectile from a height and that from a level ground.  Projectile is mostly considered as taking place along a parabolic trajectory and most candidates use the stereotype parabolic equation as against using the linear equations for vertical motion under gravity, few candidates that used the correct equations also substituted wrongly.
Time taken to reach maximum height
h = ut – ½ gt2
-50 = 30t – ½ x 10 x t2
5t2 – 30t – 50  =  0
t2 – 30 – 10  =  0
t = 6 ±
2
=   7.36 s/7.4 s
OR
V = u – gt
0 = 30 – 10t
t =   3s
Maximum height reached
V2  =  u2  -  2gx
0   =  900  -  20x
x  =  45m.
Time taken to reach the ground from maximum height
x  =  ut + ½ gt2
45 + 50 = 0 + ½  x 10t2
t  =  19   =  4.4
Time of flight  =  3 + 4.4   =  7.4s