A man standing at the top of a building, throws several balls at different angles to the horizontal. All balls are thrown at speed u = 10 m/s and it was found that all of them hit the ground making an angle of ${45}^{\circ }$ or larger than that with the horizontal. Find the height of the building (in meter). (Take g = 10 m/s²) 

Let h be the height of the building.

In y-direction, ${\mathrm{v}}_{\mathrm{y}}^2={\mathrm{u}}_{\mathrm{y}}^2+2\mathrm{gh}$            …(i)

In r-direction, ${\mathrm{v}}_{\mathrm{x}}={\mathrm{u}}_{\mathrm{x}}$                       …(ii)

Also, $\tan \mathrm{\theta }=\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}$

As per question minimum value of $\mathrm{\theta }$ is ${45}^{\circ }$.

$\Rightarrow$Ratio of minimum ${\mathrm{v}}_{\mathrm{y}}$ and maximum ${\mathrm{v}}_{\mathrm{x}}$ is $\tan {45}^{\circ }=1$

${\left({\mathrm{v}}_{\mathrm{y}}\right)}_{min}=\sqrt{2\mathrm{gh}}$ when y component of initial velocity is minimum, 

${\mathrm{u}}_{\mathrm{y}}=0\text{ and }{\left({\mathrm{v}}_{\mathrm{x}}\right)}_{max}=\mathrm{u}$ when stone is projected horizontally.

$\therefore \frac{\sqrt{2\mathrm{gh}}}{\mathrm{u}}=1\Rightarrow \frac{\sqrt{2\times 10\mathrm{h}}}{10}=1\mathrm{, }$which gives h = 5 m.