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July 29, 2014

July 29, 2014

Posted by **Noah** on Sunday, September 2, 2012 at 10:55pm.

the top of a building with an initial velocity of

11.6 m/s at an angle 52

◦

above the horizontal.

Due to gravity, the rock strikes the ground at

a horizontal distance of 19.5 m from the base

of the building.

How tall is the building? Assume the

ground is level and that the side of the building is vertical. The acceleration of gravity is

9.8 m/s

2

.

Answer in units of m

- physics -
**Elena**, Monday, September 3, 2012 at 4:36pmUpward motion of projectile to the top point

The height of this point is

hₒ= vₒ²•sin²α/2g = 11.6²•sin²52/2•9.8 = 4.26 m.

horizontal displacement is

L1 = vₒ²•sin2 α/2g=6.6 m.

Downward motion with horizontal velocity

v(x) = vₒ•cosα =11.6•cos52=7.14 m/s.

The horizontal displacement at this motion is

L2 = L-L1 =19.5-6.66=12.84 m.

The time of the motion is

t=L2/v(x) = 12.84/7.14 = 1.79 s.

The vertical displacement is

H=gt²/2 =9.8•1.79²/2 = 15.74 m.

The height of the building is

h=H-hₒ17.74-4.26 = 11.48 m.

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