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?

Answer 1

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…