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Problem :

A ball is let down a ramp on top of a table with initial velocity of 0 ms-1. When it reaches the end of the ramp, it is launched horizontally. Knowing that we don’t take air resistance or friction into account, and that the height of the ramp is h1, and that of the table is h2 relative to ground level,how do we get to the equation R (range or distance traveled from the end of the ramp) = 2√(h1h2)

Equations needed:

Since it is a projectile motion related exercise, the equations needed are Vv= vcosΘ , Vh= vsinΘ

and all the cinematic equations.

My approach

Since it’s projected horizontally i tried solving it taking into account that Vv = 0 and that Vh= t1h1. So I had to multiply gt1 by time (t2) to obtain the range. And t is equal to v/a or an alternative, t= √(2h2/g). But I wasn’t able to get to the equation R= 2√(h1h2). How do I ?

A ball is let down a ramp on top of a table with initial velocity of 0 ms-1. When it reaches the end of the ramp, it is launched horizontally. Knowing that we don’t take air resistance or friction into account, and that the height of the ramp is h1, and that of the table is h2 relative to ground level,how do we get to the equation R (range or distance traveled from the end of the ramp) = 2√(h1h2)

Equations needed:

Since it is a projectile motion related exercise, the equations needed are Vv= vcosΘ , Vh= vsinΘ

and all the cinematic equations.

My approach

Since it’s projected horizontally i tried solving it taking into account that Vv = 0 and that Vh= t1h1. So I had to multiply gt1 by time (t2) to obtain the range. And t is equal to v/a or an alternative, t= √(2h2/g). But I wasn’t able to get to the equation R= 2√(h1h2). How do I ?

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