Mayank and Akul meet an old friend
Sanchit was observing Mayank and Akul playing catch. He notices that Mayank can throw a ball, a maximum distance away from him. So if Akul stands beyond a distance , they face trouble playing the game.
One fine day it rains so heavily that Mayank and Akul are seperated by a distance of . Now in order to play, Mayank must be obviously raised to some height. They seek help from their spectator Sanchit who cleverly explains them the minimum height to which Mayank must be raised.
Can you play the role of Sanchit and help them?
If the answer is . Find .
Details and Assumptions :
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In the game of catch, two parties stand at some distance and throw a ball towards each other.
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There are no restrictions on Akul's strength
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The credit for this problem goes to Sanchit Aggarwal.
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1 solution
You should definitely nominate this for problem writing party! .
Maximum Distance a person can throw a ball = u^2 / g ,(when angle of projection is 45 degrees)
Let the person is raised by x . Assuming strength of person does not increases simply by raising HIM up :P( It requires lot of effort to do so)
Writing equations of motion in horizontal and vertical direction call speed of projection as u and angle of projection q.
3R = ucos(p)*T
x = -usin(p)*T+gT^2 / 2
Eliminate T From here to get a relation between x and Q. Differentiate both Sides and equate to zero (It would be a very elementary derivative of tanp and tan^2(p)
You will obtain tanp = 1/3 . Back substitute to get x = 4R.