A brick is thrown upward from the top of a building at an angle of 35° to the horizontal and with an initial speed of 15 m/s. If the brick is in flight for 2.7 s, how tall is the building? in (m)

Answer 1

Vo = 15m/s[35o].
Yo = 15*sin35 = 8.60 m/s.

Y = Yo + g*Tr = 0, Tr = -Yo/g = -8.60/-9.8 = 0.88 s. = Rise time.

h1 = Yo*Tr + 0.5g*Tr^2.
h1 = 8.60*0.88 - 4.9*0.88^2 = 3.77 m. Above top of bldg.

Tr+Tf = 2.7, Tf = 2.7-Tr = 2.7-0.88 = 1.82 s. = Fall time from max ht. to gnd.

h2 = 0.5g*Tf^2 = 4.9*1.82^2 = 16.2 m. Above gnd.

Hb = h2-h1 = 16.2 - 3.77 = 12.5 m = Height of the bldg.

Tr+Tf = 2.7, Tf = 2.7-Tr =

Answer 2

Oops! Disregard last line of this post.