Posted by **Big Poppa Smurf** on Thursday, November 29, 2012 at 8:45am.

A 0.54 kg rock is projected from the edge of the top of a building with an initial velocity of 13.8 m/s at an angle 59◦ above the horizontal.Due to gravity, the rock strikes the ground at a horizontal distance of 20.3 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 -
**Steve**, Thursday, November 29, 2012 at 11:45am
the vertical component of the initial velocity is 13.8 sin59° = 11.83 m/s

the horizontal component is

So, the height y is 7.11 m/s

since the rock hit 20.3 m away, it fell for 20.3/7.11 = 2.855 sec

h + 11.83*2.855 - 4.9*2.855^2 = 0

h = 6.165 m

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