Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A particle is projected from a point O with a velocity u at an angle (theta) (upwards) to the horizontal. At a certain point P it moves at right angles to its initial direction. It follows that : (A) OP makes an angle tan -1 (u/2g) to the horizontal (B) the distance of P from O is u/(2gsin(theta)) (C) the time of flight from O to P is u/(gsin(theta)) (D) the velocity of the particle at P is ucot(theta)
I understand that v is the component of instantaneous velocity of the projectile in the direction perpendicular to the initial direction of projection.then: v = (u cosθ) sinθ - (u sinθ - g t) cosθ = g t cosθ2) vx = u cosθ vy = u sinθ - g t let the direction of v = Φ. tanΦ = vy/vx = (u sinθ - g t) / (u cosθ) given vectors u and u are perpendicular. so tanΦ = - cotθ (u sin θ - g t) sinθ = - u cos² θ => u = g t sin θ Now vx = u cosθ = g t sinθ cosθ vy = gt sin² θ - g t = - g t cos²θ so v = √(vx² + vy²) = gt cosθ = u cot θ
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -