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### Q & A (Projectile Motion)

Q a projectile from a point O on the ground must hit a target P such that OP = 120 m The height of P from the ground is 60 m The minimum suitable velocity of projection is

(A ) 28 m/s (B ) 54 m/s (C ) 34 m/s ( D ) 42 m/s

IN THE SOLUTION PART THE FOLLOWING FORMULA IS USED

V min = √ g ( y + √ x ^2 + y ^2 ) AND THEY FOUND ANS ( D ) .

I DID NOT UNDERSTAND THIS FORMULA AND FROM WHERE THEY HAVE TAKEN .

Kindly help me in this regards.

Swapnil s khandekar
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The solution is attached in terms of two scanned images. Click on the images to enhance clarity.

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$