Question:
A particle is projected from the ground with an initial speed of v at an angle theta with horizontal .The average velocity of the particle between its point of projection and height of point of trajectory is
a) v/2*square root of (1+2 cos square theta) b)v/2*square root of (1+ cos square theta) c) v/2*square root of (1+3 cos square theta) d) v cos theta
Answer:
The average velocity in a uniformly accelerated motion is given by Displacement/Time.
The displacement will be: [(v2)*sin(2 ɸ)/2g]i + [(v2)*{sin(ɸ)}2/2g]j
Magnitude of displacement: (v2/2g)*[4{sin(ɸ)cos(ɸ)}2 + {sin(ɸ)}4]1/2
Time: v*sin(ɸ)/g
Average velocity: (v/2)*{4(cos(ɸ))2 + (sin(ɸ))2}, after applying given formula.
(v/2)*{3(cos(ɸ))2 + sin2(ɸ) + cos2(ɸ)},
As sin2(ɸ) + cos2(ɸ) = 1
Simplifying: (v/2)*{1 + 3(cos(ɸ))2}1/2
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