Motion in a Plane

Concept:The ball's vertical motion is independent of horizontal. The time to reach maximum height gives the initial vertical velocity, and the vertical displacement at landing time gives the building height.Formula:$v_y = u_y - g t$ and $y = u_y t - \frac{1}{2} g t^2$.Solution:Given angle $\theta = 45^\circ$, time to max height $t_{\max}=2$ s, total flight time to building $t=3$ s, $g=10\, \text{m/s}^2$.At maximum height, vertical velocity $v_y=0$. Using $v_y = u_y - g t_{\max}$: $0 = u_y - (10)(2)$ ⇒ $u_y = 20$ m/s.Now, vertical displacement when ball lands is $H$. Using $y = u_y t - \frac12 g t^2$ at $t=3$ s: $H = (20)(3) - \frac12 (10)(3)^2 = 60 - 45 = 15$ m.Answer:15 m (Option D)