Please explain how to solve the image problem fully explain clearly..........

 

Dear student

 

We know V2 = 2gh Where V is velocity of water coming out of the hole h is distance though which it has come Hence  V = √(2ghD) here h = D        (1) Also S = (1/2) gt2  ∴ t2 = (2s / g) Here s = distance travelled = H – D. Hence t2 = [{2(H – D)} / g]  i.e. t = √[{2(H – D)} / g]                (2) ∴ Horizontal distance travelled = x = V × t from (1) & (2), x = √[2gD] ×√[{2(H – D)} / g] = √[4D(H – D)] = 2√[D(H – D)] 

 

Thanks and Regards