A ball is gently dropped from a height of 2 0 m . If its velocity increases uniformly at the rate of 1 0 m/s 2 , with what velocity will it strike the ground?
Input your answer in m/s .
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As every time, excellently written :-) .
if a ball is dropped, doesn't wind resistance mean on a round object such as this its speed remains constant?
K.E = P.E
2 1 m v 2 = m g h
2 1 v 2 = g h = 2 0 0
v = 2 0 m / s
Nice Solution...
distance(s) = 20m initial velocity(u) = 0m final velocity = v acceleration = 10m/s v^2 - u^2 = 2as v^2 = 2 * 10 * 20 v^2 = 400 v = 20. Thus, final velocity is 20 m.
V^2 = 2as
V^2 = 400
V=20 m/s
Very simple just use laws of motion Apply the formulas carefully Here we apply 3rd law of motion V^2-u^2=2ag
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Here, we have the following values,
H e i g h t = s = 2 0 m
A c c e l e r a t i o n = g = 1 0 m / s 2
I n i t i a l v e l o c i t y = 0 m / s
F i n a l v e l o c i t y = ?
Substituting the values in the third equation of motion, i.e. v 2 − u 2 = 2 g s
We get, v 2 = 2 × 1 0 × 2 0
v = 4 0 0
Therfore, v = 2 0 m / s