In the figure shown, a spherical surface is fixed on ground. Its radius is $5m$. A block is placed some height below the topmost point. If acceleration of the block at any instant is $6m/s^2$, then find the height (in $m$) of the block above ground at that instant.

Find the value of $\theta$.

Given:
Radius of sphere $\left(r\right)=5m$
Acceleration of the block $\left(a\right)=6m/s^2$

From the given diagram,

$g\sin \theta =a\dots \dots \left(i\right)$

$\cos \theta =\frac{h-r}{r}\dots \dots \left(ii\right)$

By putting the values in equation $\left(i\right)$, we get,

$\sin \theta =\frac{a}{g}=\frac{6}{10}=\frac{3}{5}$

$\theta ={\sin }^{-1}\frac{3}{5}$

$\theta ={37}^{\circ }$

Find the value of $h$.

Hence, by using the value of $\theta ,r$ in equation $\left(ii\right)$ we get,

$\cos {37}^{\circ }=\frac{h-5}{5}$

As we know,

$\cos {37}^{\circ }=\frac{4}{5}$

$\frac{4}{5}=\frac{h-5}{5}$

$h=9m$

Final answer: $9m$