Change in X = 1/2at^2 a=10 t=6
1/2(10)(36)= 1/2(360)= 180m
An object is dropped from rest from the top of a 400 m cliff on Earth. If air resistance is negligible, what is the distance the object travels during the first 6s of its fall?
a)30m
b)60m
c)120m
d)180m
e)360m
I got a) as my answer. I used
x-x_i = v_i * t +1/2at
x-400= 0(6) + 1/2 (9.8)(6)
x-400=29.4
x=429.4
so then I did 429.4-400 which gave me 29.4
Is this correct?
I don't know what you are trying to do with those equations. Try using the correct formula, which is
distance fallen = (1/2) g t^2.
None of the choices is quite right but (d0 is closest. Perhaps they expect you to assume that g = 10 m/s^2. The actual value is 9.8, which you have used.
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