A body is thrown vertically upward with velocity 30 metres universe and another body is thrown up along the same vertical align with the same velocity but one second later off from the ground the two bodies meet at a height of
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11th
Physics
Motion in a Straight Line
Problems Based on Free Fall
A body P is thrown vertical...
PHYSICS
A body P is thrown vertically up with velocity 30ms
−1
and another body Q is thrown up along the same vertically line with the same velocity but 1 second later from the ground. When they meet (g=10ms
−2
)
December 26, 2019avatar
Muthu Pappachen
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ANSWER
h=ut−
2
1
gt
2
h
p
=30×1−
2
1
×10×1
2
h
p
=30−5=25m
v=u−gt
v
p
=30−10×1=20ms
−1
Let, t further time they take to meet and meet at height h
$$\begin{array}{l}h = ut - \frac{1}{2}g{t^2}\\h = {h_p} + {u_p}t - \frac{1}{2}g{t^2}\\h = {u_a}t - \frac{1}{2}g{t^2}\\{h_p} + {u_p}t - \frac{1}{2}g{t^2} = {u_a}t - \frac{1}{2}g{t^2}\\{h_p} = \left( {{u_a} - {u_p}} \right)t\\\frac{{{h_p}}}{{{u_a} - {u_p}}} = t\\\frac{{25}}{{30 - 20}} = t\\t = 2.5\,{\rm{secconds}}\\
\end{array}$$
Therefore,
P travels 1+t=3.5
Explanation:
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