# physics

A brick is thrown upward from the top of a building at an angle of 40° to the horizontal and with an initial speed of 16 m/s. If the brick is in flight for 3.4 s, how tall is the building?

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1. At time t after throwing,

Y (measured from the ground) = H + 16sin40*t - 4.9 t^2

H is the building height.

Set Y = 0 at t = 3.4s and solve for H

H = 0 + 4.9*(3.6)^2 - 34.97 = 28.53 m

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posted by drwls

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