Questions
A particle is projected from the ground with an initial speed of v at an angle with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is:
detailed solution
Correct option is C
Average velocity = DisplacementTime vav = H2+R24T2----(i)Here, H = maximum height = v2sin2θ2gR = range = v2sin2θgand T = time of flight = 2v sinθgvav = v21+3 cos2θTalk to our academic expert!
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A ball is projected obliquely with a velocity 49 ms–1 strikes the ground at a distance of 245 m from the point of projection. It remained in air for
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