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 .

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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! 
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Draw a square ABCD.
A       B
C       D
Join 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/x
Or,x=h
h=uyt+(1/2)gt2
h=gt2/2
Also x=uxt=vt
From h=x,
gt2/2=vt
or,gt/2=v
Or,t=2v/g
So x=vt=2v2/g
Hope this helps
						

							
So it touches the plane from the point of projection at a distance
This is the answer.Hope this helps.Thanks for disapproving.