Physics, asked by cuteemano956gmailcom, 1 year ago

plz tell me how to solve part 1 .
time for max horizontal distance...

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Answered by vvipjattin

0

t=2u sin Q/g were Q is tita and range = u^2sing2Q/g

Answered by Sumitmbbs

0

For maximum horizontal distance, angle of projection must be 45 degree with the horizontal.

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