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2.Motion in Straight Line
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A man drops a ball downside from the roof of a tower of height $400 m\,eters$. At the same time another ball is thrown upside with a velocity $50\, meter/sec$. from the surface of the tower, then they will meet at which height from the surface of the tower............$meters$
A
$100 $
B
$320$
C
$80$
D
$240 $
Solution

(c) Let both balls meet at point $P $ after time $t$.
The distance travelled by ball $A$, ${h_1} = \frac{1}{2}g{t^2}$
The distance travelled by ball $B$, ${h_2} = ut – \frac{1}{2}g{t^2}$
${h_1} + {h_2} = 400\;m$
$⇒$ $ut = 400,\;t = 400/50 = 8\;sec$
$\therefore {h_1} = 320\;m\;\;{\rm{and}}\;\;{h_2} = 80\;m$
Standard 11
Physics
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