Question

A marble is thrown horizontally with a speed of 10.2 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 76 ° with the horizontal. From what height above the ground was the marble thrown?

Answer 1

Let mahnitide of final velocity be V.

Its Horizonta component = V cos 76

But horizontal component doesn't change.

so,

V cos 76 = 10.2

V= 10.2/cos 76 = 42.2 m/s

Vertical component of final velocity= V* sin 76 = 42.2 sin 76 =40.95 m/s

For vertical motion use:

Vf ^2= Vi^2 + 2*g*h

40.95^2 = 0^2 +2*9.8*h

h=85.5 m

Answer: 85.5 m