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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 .


2 years ago

## Answers : (4)

Tony
108 Points
							You

2 years 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!

2 years 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

2 years 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.

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