A body P is thrown vertically up with velocity 30ms^-2 and another body Q is thrown up along the same vertically line with same velocity but 1 second later from the ground. When they meet (g=10ms^-2)
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1
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
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