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08/01/2021 11:21 am
Topic starter
A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
1 Answer
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08/01/2021 11:27 am
Here, time of ascent = time of descent,
\(t = \frac{6}{2}\) = 3s
(a) v = 0, a = -g = -9.8 m/s2
From v = u + gt
0 = u - 9.8 x 3
u = 29.4 m/s
(b) From v2 - u2 = 2gs
0 - (29.4)2 = 2(-99.8)h
\(\rightarrow h = \frac{29.4 \times 29.4}{2 \times 9.8}\) = 44.1 m
(c) t = 3s, ball is at maximum height.
From, \(s = ut + \frac{1}{2}gt^2\)
h = \(0 + \frac{1}{2} \times 9.8(1)^2\)
= 4.9 m below the top.