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        a ball is projected horizontal with a speed v from the top of a plane inclined at an angle 45 degrees with the horizontal . how far from the point of projection with the ball strike the plane .
one year ago

Tony
108 Points
							You

one year ago
Tony
108 Points
							You can easily solve all the questions of horizontal projectile from an inclined plane by the following formulas :Let assume that angle of inclination is A.Velocity with which it is projected is u m/s and g is acaccelerati due to gravity!Time (T) = 2u sinA / g cos A                =2u tanA / g Distance along the inclined plane ( that is, the distance at which projected body strikes the plane) = (u cos A)×T + 1/2 ×g sin A ×T^2 Distance perpendicular to plane=u sinA ×T - 1/2 g cos A ×T^2 Hope it helps you! Plzz approve it if you like the answer!

one year ago
Susmita
425 Points
							Draw a square ABCD.A       BC       DJoin A,D.The angle ADC is 45°.The particle is projected horizontally.So there is no vertical component of velocity.ux=v and uy=0.There is no horizontal component of acceleration but there is vertical component due to Earth's gravity.ax=0 and ay=g.Let AC=h and CD=x.We have to find out x.tan45=h/xOr,x=hh=uyt+(1/2)gt2h=gt2/2Also x=uxt=vtFrom h=x,gt2/2=vtor,gt/2=vOr,t=2v/gSo x=vt=2v2/gHope this helps

one year ago
Susmita
425 Points
							So it touches the plane from the point of projection at a distance$AD=\sqrt{x^2 +h^2} =\sqrt{2.(2v^2 /g)^2}=\sqrt{2}(2v^2/g)$This is the answer.Hope this helps.Thanks for disapproving.

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions