A particle starts from the origin at t=0 with a velocity of 8.0jm/s and moves in the x-y plane with a constant acceleration of (4.0i^+2.0j^)m/s2. At the instant the particle's x-coordinate is29 m, what is its y-coordinate?
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
88.9 m
b
44.9 m
c
84.9 m
d
54.9 m
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
x=uxx+12axt2 ∴ 29=(0)(t)+12×(4.0)t2 t2=14.5 s2 t=3.8 s y=uyt+12ayt2 =(8)(3.8)+12×2×14.5 =44.9 m