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svj29

Q. There is a road. Its two ends are separated by a distance of 22m.

On one end of the road there is a building of height x metres.

and exactly opposite to the building, there is another building of height

x+9 metres. A ball is thrown from the taller building at the speed of 2m/s

making an angle of 45 degree with the horizontal and simultaneously another ball (same to the previous one) is thrown from the shorter building at a speed of 14m/s

making an angle of 45 degree with the horizontal.

Calculate the time when they will the closest. and the distance b/w them

when they will be nearest to each other.

I DON'T KNOW THE ANSWER...PLZ SOLVE AND TELL THE ANSWER.

5 SALUTES ASSURED TO THE CORRECT ANSWER

bsgdabest

x coordinate of 1st particle: 2cos45*t
y "                     "            : H-(x+9)=2sin45-1/2gt^2 ( kinda like translation of axes)
similiarly X coordinate of 2nd particle: 22-(14cos45*t)

             Y "                  "              : x+ 14sin45*t - 1/2gt^2
posn vector of the first stone: (t*sqrt2 , x+9+t*sqrt2 -5t^2)
                           second stone: ( 22-7t*sqrt2,x+7t*sqrt2-5t^2)
distance between those 2 pts=d^2= f(t)=200t^2 - tsqrt2*460 + 565

min. at t= 1.15sqrt2

u can substitute value of t in d^2 to find that value too

anchitsaini

taking origin to be the place from where ball is thrown

with 14m/s--

after time t (for 14m/s ball)--

coordinates would be
s1= [14cos45 t] i + [14sin45 t - 5t^2]j

similarly for ball thrown with 2m/s velocity--

s2= [22 - 2cos 45 t] i + [9 -{2sin45 t + 5t^2}] j

s=s2-s1=[22- 16cos 45 t]i + [9 - 16 sin 45 t]j

=root[256 t^2 - 701.45t + 565]

s would be minimum when ds/dt=0

or when

(512t - 701.45)/denominator=0

or t=701.45/512
= 1.37s

anchitsaini

figure

sinjan.j

here are just the steps

1) Write down the equations of projectile for both the balls.
2) Find x1(t), y1(t) and x2(t), y2(t).... coordinates of both the balls as a function of time.
3) write the expression of distance between the two balls as a function of time now.
4) Differentiate and put to zero (minimize distance)

am I correct?????

sinjan.j

( take taller building on left, shorter on right, foot of taller building =(0,0), foot of shorter one is (22,0)

x coordinate of 1st particle: 2cos45*t
y " " : H-(x+9)=2sin45-1/2gt^2 ( kinda like translation of axes)
similiarly X coordinate of 2nd particle: 22-(14cos45*t)

Y " " : x+ 14sin45*t - 1/2gt^2
posn vector of the first stone: (t*sqrt2 , x+9+t*sqrt2 -5t^2)
second stone: ( 22-7t*sqrt2,x+7t*sqrt2-5t^2)
distance between those 2 pts=d^2= f(t)=200t^2 - tsqrt2*460 + 565

min. at t= 1.15sqrt2

u can substitute value of t in d^2 to find that value too

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