Questions
A projectile is projected with initial velocity , then what is the horizontal range of the projectile?
a
8.8 m
b
13.5 m
c
9.6 m
d
11.2 m
detailed solution
Correct option is C
Initial velocity is (6i^+8j^)m/s (given).Magnitude of velocity of projection isu=ux2+uy2=62+82=10m/sAngle of projectiontanθ=uyux=86⇒sinθ=45 and cosθ=35Now the horizontal range is,R=u2sin2θg=u22sinθcosθg=(10)2×2×(4/5)×(3/5)10=9.6mTalk to our academic expert!
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