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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

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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Hint: The detailed proof can be seen in this post. Then proceed as follows:

We have, ${u_{\min }} = \sqrt {g(r + h)} $

${u_{\min }} = \sqrt {9.8 \times (120 + 60)} $

${u_{\min }} = \sqrt {1764}  = 42m/s$