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 16 m/s. If the brick is in flight for 3.1 s, how tall is the building?

Answer 1

The vertical component of the speed is initially
Voy = 16 sin25 = 6.762 m/s.

Height above ground is

y = H + Voy*t - (g/2)t^2

y = H +6.762t -4.9 t^2

You know that y = 0 when t = 3.1 s.

Solve for H, the building height.