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 )
Live 1-1 coding classes to unleash the creator in your Child
A particle is projected with speed u so as to strike at right angles a plane through the point of projection inclined at 30° with the horizon. The range on this inclined plane is a) 4u^2/7g b) 3u^2/5g c) u^2/2g d) u^2/g
Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u . The point O is called the point of projection, θ is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight. Now, (a). The total time of flight is Resultant displacement is zero in Vertical direction. Therefore, by using equation of motion s=ut− 21 gt 2 gt=2sinθ t= g 2sinθ ?(b). The horizontal range is Horizontal range OA = horizontal component of velocity × total flight time R=ucosθ× g2usinθ ?R= gu2 sin2θ ?(c). The maximum height is It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero. By using equation of motion v 2 =u 2 −2as 0=u 2 sin 2 θ−2gH H= 2g u 2 sin 2θ? Hence, this is the required solution
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -