A brick is thrown upward from the optimal of a building at an angle of 25˚ come the horizontal and with one initial rate of 15 m/s. If the brick is in flight for 3.0 s, how tall is the building?
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You are watching: A brick is thrown upward from the top of a building at an angle of let us job the position of the brick to the vertical axis. Signify the initial height by H (it is the height of the building) and also obtain the complying with equation:

H(t) = H + V0*sin(a)*t - (g*t^2)/2.

here V0 is the early velocity (15m/s), a is one angle above horizontal...

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Let us task the position of the brick to the vertical axis. Signify the initial height by H (it is the height of the building) and also obtain the complying with equation:

H(t) = H + V0*sin(a)*t - (g*t^2)/2.

Here V0 is the early stage velocity (15m/s), a is an angle over horizontal (25°) (+ is indigenous the "upward"). And also g is the acceleration of heaviness (approx. 9.8m/s^2).

We understand that H(3s)=0, so

H = g*(3^2)/2 - V0*sin(a)*3 = 9.8*4.5 - 15*sin(25°)*3.

Use calculator and obtain the answer, 44.1-19=25.1(m).

Of course we disregard the air resistance here.

See more: What Is 189 Cm In Feet And Inches ? What Is 189 Cm In Feet And Inches

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