# 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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### 2 Answers

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.

This problem can be solved by the standard equation

`S = (U)*t +(1/2)*a*(t^2)`

so , going to the question , Given that

angle of projection of the brick =** 25 deg**

initial velocity (U) = **15 m/s**

total time brick is in flight= **3.0 s **

The brick is projected upwards it is projected as shown in the attachment

Let S be the height of the building

and the formula is modified as

`S = (U_y)*t - (1/2)*g*(t^2)`

as the brick is thrown against gravity so the acceleration becomes a = (-g)

velocity towards `U_y` = 15* (sin 25) = 6.34 m/s

so,

`S = (U_y)*t - (1/2)*g*(t^2)`

= (6.34)*(3) - (1/2)*(9.8)(3^2)

= -25.1 m

As the height cannot be negative so **S = 25.1 m**