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# Answer to Question #705 in Mechanics | Relativity for brian

Question #705
A 0.49kg rock is projected from the edge of the top of a building with an initial velocity of 8.36 m/s at an angle 60◦ above the horizontal.Due to gravity, the rock strikes the ground at a horizontal distance of 10.7 m from the base of the building.How tall is the building?

The acceleration of gravity is
9.8 m/s2 .
Expert's answer
A 0.49kg rock is projected from the edge of the top of a building with an initial velocity of 8.36 m/s at an angle 600; above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 10.7 m from the base of the building. How tall is the building?

vy0 = v0 * sin60&deg; = 8.36 * &radic;3/2 = 7.24 m/s
vx0 = v0 * cos60&deg; = 8.36 * 1/2 = 4.18 m/s

Total time of flight:

t = d &divide; vx0 = 10.7m &divide; 4.18 m/s = 2.56 s

Time of flight over the top of the building level:

t&prime; = 2t_peak = 2 * vy0 &divide; g = 2 * 7.24 m/s &divide; 9.8 m/s&sup2; = 1.48 s

Hence, height of the building is:

h = vy0 * (t - t&prime;) + g * (t - t&prime;)&sup2; &divide; 2 = 7.24 m/s * (2.56 s - 1.48 s) + 9.8 m/s&sup2; * (2.56 s - 1.48 s)&sup2; &divide; 2 = 13.53 m

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