Posted by **Anonymous** on Sunday, June 10, 2012 at 10:13pm.

A marble is thrown horizontally with a speed of 10.5 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 34.1 ° with the horizontal. From what height above the ground was the marble thrown?

- physics -
**Elena**, Monday, June 11, 2012 at 4:00am
tan φ=v(y)/v(x) ,

v(y)=v(x)• tanφ.

v(y)=g•t, => t=v(y)/g

h=gt²/2 =g• v²(y)/2•g²

= v²(y)/2•g= (v(x)• tanφ)²/ 2•g= (10.5•tan 34.1º)²/2•9.8 =2.6 m.

The answer is correct, but the given data are strange enough , especially, the angle…

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