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THE SPEED OF THE PROJECTILE WHEN IT IS AT THE MAXIMUM HEIGHT IS ROOT 2/5 TIMES OF ITS SPEED WHEN IT IS AT HALF OF THE MAXIMUM HEIGHT THE ANGLE OF PROJECTION IS?

THE SPEED OF THE PROJECTILE WHEN IT IS AT THE MAXIMUM HEIGHT IS ROOT 2/5 TIMES OF ITS SPEED
WHEN IT IS AT HALF OF THE MAXIMUM HEIGHT THE ANGLE OF PROJECTION IS? 

Grade:11

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
Speed of shot at the most astounding point = ucosα (α is the edge of projection) 
stature of the projectile= u2sin2α/2g 
MAX ht/2= u2sin2α/4g 
Speed at MAX ht/2 = v 
u2-v2=2gh 
=>v2=u2-2gx u2sin2α/4g 
=>v2=u2(1-sin2α/2) 
ucosα=√(2/5) √u^2 (1-sin^2⁡α/2) 
=>cosα=√(2/5) √(1-sin^2⁡α/2) 
=>cos2α=(2-sin2α)/5 
=>5cos2α=1+cos2α 
=>cos2α=1/4 
=>α=60⁰ .
So the prjectijon of angle found is 60 degree.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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