Physics
posted by Big Poppa Smurf .
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

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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