According to question, ball reaches at top of a
tower at the time, $t_1=3 \mathrm{~s}$ and further reaches at the maximum height at time, $t_2=2 \mathrm{~s}$.
$\therefore$ Total time taken by ball to reached the maximum height,
$\begin{array}{rlrl} & & t & =t_1+t_2=3+2 \\ \therefore & & t=5 \mathrm{~s}\end{array}$
If $u$ be velocity of ball at time, $t=0$, then from the first equation of the motion,
$\begin{aligned} & v=u-g t \quad\left[\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right] \\ & 0=u-10 \times 5 \Rightarrow u=50 \mathrm{~m} / \mathrm{s}\end{aligned}$
If $h$ be the height of the tower, then from the second equation of the motion,
$\begin{aligned} h=u t-\frac{1}{2} \times g t^2 & =50 \times 3-\frac{1}{2} \times 10 \times 3^2 \\ & =150-45=105 \mathrm{~m}\end{aligned}$
Hence, the height of the tower is $105 \mathrm{~m}$.