Questions
A particle is projected with a velocity m/s. The time after which it will start moving perpendicular to its initial direction of motion is
detailed solution
Correct option is D
Given v1 → = 8i^+6j^ m/sAcceleration due to gravity, a→ = -gj^ = -10j^ m/s2Velocity after time t; v1 →. v2→ = 08i^+6j^.8i^+6-10tj^ = 064+6(6-10t) = 064 + 36 - 60t = 060t = 100 t = 53 secTalk to our academic expert!
Similar Questions
A ball is projected obliquely with a velocity 49 ms–1 strikes the ground at a distance of 245 m from the point of projection. It remained in air for
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
