Question

A marble is thrown horizontally with a speed of 22.1 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 46 ° with the horizontal. From what height above the ground was the marble thrown?

Answer #1

tan angle = Vfv/Vfh

Vfv is final vertical velocity

Vfh is final horizontal velocity which is same as initial horizontal velocity since horizontal velocity do not change

tan angle = Vfv/Vfh

tan (46) = Vfv / 22.1

1.036 = Vfv / 22.1

Vfv = 22.9 m/s

now use for vertical direction:

vi = 0

vf = -22.9 m/s (negative because it is in downward direction)

a = -9.8 m/s^2

use:

vf^2 = vi^2 + 2*a*d

(-22.9)^2 = 0^2 + 2*(-9.8)*d

524.4 = -19.6*d

d = - 26.8 m

This is negative because distance moved is in downward direction

So, height = 26.8 m

Answer: 26.8 m

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