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Question
A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take g = 10 m/s2.
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Solution
Given:
Distance travelled by the ball in 0.200 seconds = 6 m
Let:
Time, t = 0.200 s
Distance, s = 6 m
a = g = 10 m/s2 (Acceleration due to gravity)
Using the equation of motion, we get:
\[s = ut + \frac{1}{2}a t^2 \]
\[6 = u\left( 0 . 2 \right) + \frac{1}{2} \times 10 \times 0 . 04\]
\[ \Rightarrow u = \frac{5 . 8}{0 . 2} = 29 \text{ m } /s\]
Let h be the height from which the ball is dropped.
We have:
u = 0 and v = 29 m/s
Now,
\[h = \frac{v^2 - u^2}{2a}\]
\[\Rightarrow h = \frac{{29}^2 - 0^2}{2 \times 10} = \frac{29 \times 29}{20} = 42 . 05 \text{ m } \]
∴ Total height = 42.05 + 6 = 48.05 m ≈ 48 m
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