Physics - Kinematics Question with Solution | TestHub

Let $t$ be the total time of fall.

The distance traveled in $t$ seconds is $h=\frac{1}{2}g{t}^2$.

The distance traveled in $(t-1)$ seconds is $h^{\prime }=\frac{1}{2}g(t-1{)}^2$.

The distance in the last second is $h-h^{\prime }=24.5$ m.

So, $\frac{1}{2}g{t}^2-\frac{1}{2}g(t-1{)}^2=24.5$.

$\frac{1}{2}g(t^2-(t^2-2t+1))=24.5$.

$\frac{1}{2}g(2t-1)=24.5$.

Using $g=9.8$ m/s², we get $4.9(2t-1)=24.5$.

$2t-1=5$, so $2t=6$, and $t=3$ s.

The height of the tower is $h=\frac{1}{2}g{t}^2=\frac{1}{2}(9.8)(3^2)=4.9\times 9=44.1$ m.