Kinematics Question 662

Question: The angle which the velocity vector of a projectile thrown with a velocity v at an angle θ to the horizontal will make with the horizontal after time t of it’s being thrown up it’s:

Options:

A) θ

B) tan1(θ/t)

C) tan1(v cosθv sinθgt)

D) tan1(v sinθgtv cosθ)

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Answer:

Correct Answer: D

Solution:

[d] Horizontally after time vcosθ=vcosβ -(i) [β = angle with horizontal after time t]

Vertically, vsinθgt=vsinβ -(ii)

Dividing on (ii)/ (i) we get tanβ=vsinθgtvcosθ

tanβ=vsinθgtvcosθβ=tan1(vsinθgtvcosθ)



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