My 100 followers question

Two small stones A , B A ,B of masses 1 k g 1~kg and 2 k g 2~kg respectively, are projected simultaneously at t = 0 t=0 . The stone A A is projected horizontally with speed V a m s 1 V_a~ms^{-1} from the top of a tower that is Y Y meters high and the stone B B is projected from the ground at a distance X X meters from the foot of the tower with speed V b = 90 m s 1 V_b~=~90~ms^{-1} at an angle of 3 7 o 37^o with the horizontal as shown in the figure. At time t = 6 s e c t~=~6~sec after projection, they collide in air and at that moment their velocities are mutually perpendicular. Then find X + Y + V a X~+~Y~+~V_a

Note: give the sum of the numerical values of X , Y , V a X, Y , V_a .


The answer is 791.000.

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1 solution

Aryan Goyat
Jul 11, 2015

write the velocity in vector form Va=v(-i)+gt(-j)-------1 Vb=90cos37 (i)+90sin37-gt(j)-------2 at 6s they are perpendicular =-->Va.Vb=0(dot product)----3 putting t=6 and combining 1,2,3- Va=5 x=(90cos37+5) 6=462 y(sitting in frame of b)=90sin37*6=324 -->x+y+Va=791

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