Hard
Question
A projectile is thrown with a speed v at an angle θ with the vertical. Its average velocity between the instants it crosses half the maximum height is
Options
A.v sin θ, horizontal and in the plane of projection
B.v cos θ, horizontal and in the plane of projection
C.2v sin θ, horizontal and perpendicular to the plane of projection
D.2v cos θ, vertical and in the plane of projection.
Solution
Avg. vel. b/w A & B
( Acceleration is constant = g)Now if
= V1xî + V1yĵThen
= V1x î − V1y ĵ (∵ both A & B are at same lavel)∴
= V1x = V sin θ (θ is from vertical θ) Create a free account to view solution
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