A brick is thrown upward from the top of a building at an angle of 250 to the ho

Question

A brick is thrown upward from the top of a building at an angle of 250 to the horizontal and with an initial speed of 15 m/s.

a. If the brick is in flight for 3.0 s, how tall is the building?

b. With what speed and angle of impact does the brick land?

c. Sketch the graphs of vs. time and vs. time for the brick from the time it is thrown to the time it hits the ground.

d. Suppose that a second brick is thrown upward with an angle of 400 to the horizontal and with an initial speed of 15 m/s. How much longer will the second brick remain in the air compared to the first brick?

Explanation / Answer

A) let h be the height of the building

-h = (uy*t)-(0.5*g*t^2)

-h = (15*sin(25)*3)-(0.5*9.8*3*3) = -25.08

height of the tower h = 25 m

B) vx = 15*cos(25) = 13.6 m/s

vy = uy-(g*t) = (15*sin(25))-(9.8*3) = -23 m/s

v = sqrt(13.6^2+23^2) = 26.77 m/s

theta = atan(-23/13.6) = 59.4 degrees below the horizontal

C) question is not clearly asked

D) let -h = (uy*t)-(0.5*g*t^2)

uy = 15*sin(40) = 9.64 m/s

-25 = (9.64*t) -(0.5*9.8*t^2)

4.9*t^2 -(9.64*t) - 25 = 0

solving we get t = 3.44 s

So 0.44s more time second brick will stay in air compared with the first brick

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