Kinematics Question 662
Question: The angle which the velocity vector of a projectile thrown with a velocity v at an angle $ \theta $ to the horizontal will make with the horizontal after time t of it’s being thrown up it’s:
Options:
A) $ \theta $
B) $ {{\tan }^{-1}}( \theta /\text{t} ) $
C) $ {{\tan }^{-1}}( \frac{\text{v cos}\theta }{\text{v sin}\theta -\text{gt}} ) $
D) $ {{\tan }^{-1}}( \frac{\text{v sin}\theta -\text{gt}}{\text{v cos}\theta } ) $
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Answer:
Correct Answer: D
Solution:
[d] Horizontally after time $ v\cos \theta =v\cos \beta $ -(i) [$ \beta $ = angle with horizontal after time t]
Vertically, $ v\sin \theta -gt=v\sin \beta $ -(ii)
Dividing on (ii)/ (i) we get $ \tan \beta =\frac{v\sin \theta -\text{gt}}{v\cos \theta } $
$ \tan \beta =\frac{v\sin \theta -\text{gt}}{v\cos \theta }\Rightarrow \beta ={{\tan }^{-1}}( \frac{v\sin \theta -gt}{v\cos \theta } ) $
BETA
