Questions
For the projectile shown in figure, the time of flight and range along the inclined plane is: (g=10 m/s2)
detailed solution
Correct option is B
time of flight=T=2usinα+βgcosβ T=(2)(202)sin(450+300)(10)cos300 T=(2)(202)sin(750)(10)32=8×1.414×0.963=6.31s range=R=u2gcos2βsin2α+β+sinβ R =(202)2[sin(2×450+300)+sin300](10)cos2300=145.71mTalk to our academic expert!
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