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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.  ### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$