Question

help me to get rid from this question​

Answer 1

Look u need to calculate the motion separately for x and y directions.initial velocity in y dirc. is 6 acc. is -3.So max. diatance covered in y direction is till the particle starts retarding, So

0=6^2-2*3*Sy

Sy=6

Now in x direction there is no initial velocity but there is acc. of 2 So

Sx=0+1/2*2*t^2......(1)

Now t will be the time when y coordinate bcms max as asked in question.so

Vy=Uy+at

0=6-3t >t=2

plcing t in (1)> Sx=0+1/2*2*4=4

So the coordinates are (4,6).