Grade 11Mechanics

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 .

Khushi Shahi

7 Years agoGrade 11

4 Answers

Tony

7 Years ago

You

Tony

7 Years ago

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!

Susmita

7 Years ago

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.

u_x=v and u_y=0.

There is no horizontal component of acceleration but there is vertical component due to Earth's gravity.

a_x=0 and a_y=g.

Let AC=h and CD=x.We have to find out x.

tan45=h/x

Or,x=h

h=u_yt+(1/2)gt²

h=gt²/2

Also x=u_xt=vt

From h=x,

gt²/2=vt

or,gt/2=v

Or,t=2v/g

So x=vt=2v²/g

Hope this helps

Susmita

Approved Tutor Answer7 Years ago

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.

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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 .

Khushi Shahi

4 Answers

Tony

Tony

Susmita

Susmita

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