A basejumper jumping off a cliff initially undergoes free fall with negligible a
ID: 1538324 • Letter: A
Question
Explanation / Answer
i)from eqn of motion
S = ut + 1/2 at^2
In our case, S = H ; t = 1 , a = g
H = 1/2 x 9.8 x 1 = 4.9 m
H1 = 4.9 m
ii)from t = 1 to 2
H2 = 1/2 x 9.8 x 1 = 4.9
H2 = 4.9 m
v2 =
total fall = 4.9 + 4.9 = 9.8 m
iii)t = 2 to 3
H3 = 4.9 m
total fall = 4.9 + 4.9 + 4.9 = 14.7 m
iv)H4 = 4.9 m
total fall = 4.9 +4.9 + 4.9 + 4.9 = 19.6 m
H4 = 19.6 m
b)v = u + at
v = 9.8 x 4 = 39.2 m/s
Hence, v = 39.2 m/s
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