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?
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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