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**silvercats** Ball is projected in a 30m/s velocity with a angle of X from a place called 'O' horizontally.The ball was caught by someone at 1m of height .This is position 'P'.If tan X=4/3 ,what is the time took for the ball to go to 'P' from 'O' ?what is the displacement of the ball at the 'P' position?

I never got the book's answer.here is how I have done it incorrectly :

Y component of 30m/s is 24m/s,X is 18(took from tan x=4/3)

to find the time \uparrow OP s=ut+1/2atē

1=24t+1/2(-10)tē

5tē=24t

t=24/5 <<which is wrong

finding OP distance \to s=ut

s=18*24/5

s=86.5 <<which is wrong

what are the correct ones?

0.447s,1.28m are the answers give in the book.

thanks