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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)
