A brick is thrown upward from the top of a building at an angle of 25˚ to the horizontal and with an initial speed of 15 m/s. If the brick is in flight for 3.0 s, how tall is the building?
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Let us project the position of the brick to the vertical axis. Denote the initial height by H (it is the height of the building) and obtain the following equation:

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

Here V0 is the initial velocity (15m/s), a is an angle above horizontal...


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Let us project the position of the brick to the vertical axis. Denote the initial height by H (it is the height of the building) and obtain the following equation:

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

Here V0 is the initial velocity (15m/s), a is an angle above horizontal (25°) (+ is from the "upward"). And g is the acceleration of gravity (approx. 9.8m/s^2).

We know 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 neglect the air resistance here.

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