A body is project at t= 0 with a velocity `10 ms^-1` at an angle of `60 ^(@)` with the horizontal .The radius of curvature of its trajectory at t=1s is R. Neglecting air resitance and taking acceleration due to gravity `g= 10 ms^-2`, the value of R is :
A
10.3 m
B
2.5m
C
5.1
D
2.8m
Text Solution
Verified by Experts
The correct Answer is:
D
t=1s , Vx=5m/s `Vy=5sqrt3-10xx1` `V^z/R_c=g cos alpha ` i.e., `R_C=V^2/(g sin alpha) = 26.8/(10 cos alpha)=(26.8sqrt(26.5))/50` After calculations , we get `R_C` 2.7 m Closest option is 2.8 m
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VMC MODULES ENGLISH-JEE Main Revision Test-6 | JEE-2020 -PHYSICS
