Journey in a train is adventurous particularly when you have a seat. The girl sitting near window ate a banana and dropped the peel from the window. Her co-passenger looking through the window found that it dropped vertically down and touched the ground in $0.2 s$ . After sometime she requested her sister sitting on the upper berth to drop a chocolate bar.The sister dropped the bar, but it fell in front of the girl instead of reaching her hand. She was angry but the co-passenger calmed her by saying that she dropped exactly in line of your hand but as the train is accelerating it did not reach you and fell in front of you.
Two particles are thrown with $8 m s^{- 1}$ as shown in the figure, one horizontally from a height of $40 m$ ad the other from a height of $50$ in making an angle of $60^{\circ}$ with the vertical. They strike in mid air. Find the coordinates of strike point and distance between the buildings.

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Solution

$v_{A} = 8^i v_{B} = - 4 \sqrt{3}^i - 4^j v_{B / A} = (- 4 \sqrt{3} - 8)^i - 4^j t h i s v e l o c i t y s h o u l d t b e a l o n g t h e i n i t i a l l i n e j o i n i n g t h e t w o p o i n t s t a n θ = \frac{4}{(4 \sqrt{3} + 8)} = \frac{50 - 40}{x} \Rightarrow x = 10 (\sqrt{3} + 2) t i m e f o r c o l l i s i o n = \frac{x}{v_{r e l a l n o g x a x i s}} = \frac{10 (\sqrt{3} + 2)}{4 \sqrt{3} + 8} = 2.5 s x c o - o r d i n a t e = 8 \times 2.5 = 20 m y c o - o r d i n a t e = \frac{1}{2} g t^{2} = 31.25 m$

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vA=8^ivB=−4√3^i−4^jvB/A=(−4√3−8)^i−4^jthisvelocityshouldtbealongtheinitiallinejoiningthetwopointstanθ=4(4√3+8)=50−40x⇒x=10(√3+2)timeforcollision=xvrelalnogxaxis=10(√3+2)4√3+8=2.5sxco−ordinate=8×2.5=20myco−ordinate=12gt2=31.25m