Let [tex]y_0[/tex] be the height of the building and thus the initial height of the ball. The ball's altitude at time [tex]t[/tex] is given by
[tex]y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.
The ball reaches the ground when [tex]y=0[/tex] after [tex]t=2.9\,\mathrm s[/tex]. Solve for [tex]y_0[/tex]:
[tex]0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2[/tex]
[tex]\implies y_0\approx16.258\,\mathrm m[/tex]
so the building is about 16 m tall (keeping track of significant digits).
