A boy throws a ball into air at 45° from the horizontal to land it on a roof of a building of height H . If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is ___ m. $$(g=10m/s^{2})$$

We need to find the height $$H$$ of the building where a ball reaches its maximum height in 2 s and lands on the building in 3 s.

Since the ball is projected at $$45°$$, its horizontal and vertical components of the initial velocity are equal, namely $$u_x = u_y = \frac{u}{\sqrt{2}}$$. The time to reach the maximum height is $$t_{max} = \frac{u_y}{g} = \frac{u/\sqrt{2}}{10} = 2$$, which gives $$u = 20\sqrt{2}$$ m/s and therefore $$u_x = 20$$ m/s and $$u_y = 20$$ m/s.

To find the height after 3 s, we use the vertical displacement formula $$H = u_y t - \frac{1}{2}gt^2$$. Substituting $$u_y = 20$$ m/s and $$t = 3$$ s into this expression yields $$H = 20(3) - \frac{1}{2}(10)(9) = 60 - 45 = 15$$ m.

Hence, the height of the building is $$H = 15$$ m. Therefore, the answer is Option D.