Given that the initial(at #t=0#) velocity of projection of the particle is #u =40"m/s"# in the direction making an angle #theta =53^@# (upward) with the horizontal.
Initially at position O
# t = 0#
The horizontal component of velocity #u_x=ucostheta#. This will remain unaltered throughout its movement as it is orthogonal to gravitational force.
The vertical component of velocity #u_y=usintheta# . This will change as the force of gravitation acts opposite to it. .
Let at position P the velocity of the particle becomes perpendicular to the initial velocity #u#. So at this position the velocity of the particle will subtend angle #=90^@-theta# (downward) with the horizontal
At position P
The horizontal component of velocity #u_x=ucostheta#.
The actual velocity #v# at this position is inclined with the constant horizontal component at an angle #90-theta#.
So
#=>vcos(90-theta)=ucostheta#
#=>vsintheta=ucostheta#
#=>v=(ucostheta)/sintheta=ucottheta#
Now acceleration due to gravity #g # is always orthogonal to horizontal component of velocity but it is inclined with the velocity #v # at position P at an angle #theta # . So the component of acceleration due to gravity orthogonal to v is #a=gsintheta#
This acceleration (a) being orthogonal with v can be treated as centripetal acceleration and its relation with radius of curvature #r# is as follows
#a=v^2/r#
#=>gsintheta=(u^2cot^2theta)/r#
#=>r=(u^2cot^2theta)/(gsintheta)#
Inserting #u=40"m/s"; theta =53^@;g=10"m/"s^2#
#cot53^@~~0.75 and sin53^@~~0.8#
we get
Radius of curvature at P
#r=(40^2cot^2 53)/(10xxsin53)=112.5m# which is option (3)
