1ur differentiation is wrong
also i canlt understand ur solution clearly
how y=bt-ct2
explain in detail please taking in account the acc in diff directions
A particle is projected from a horizontal plane(x-z plane) such that its velocity vector at time t is given by V=ai+(b-ct)j . Its range on horizontal plane is
a) ba/c b)2ba/c c)3ba/c d)None
-
UP 0 DOWN 0 0 4
4 Answers
1Anyone please try this.
The answer given is b
I need explanation how to solve this
1
3its b my dear
range =vx*T
now in y axis, acceleration =-cj^
and initial velocity in y direction initial velocity =b
thus t=-u/a=b/c
T=2t=2b/c
range =2ab/c as vx=a