Questions
A particle is projected with a velocity , so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is
detailed solution
Correct option is A
H=u2sin2θ2g and R=u2sin2θg Since,R=2H, so u2sin2θg=2u2sin2θ2g or 2sinθcosθ=sin2θ or tanθ=2 ∴R=u22gsinθcosθ=2u2g2515=4u25g.Talk to our academic expert!
Similar Questions
A ball is thrown with a velocity of u making an angle with the horizontal. Its velocity vector normal to initial vector (u) after a time interval of
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
