An object is dropped freely under gravity from a height h. If during the last second of its motion it travels 36% of the total distance, what is the value of h?

What the “last second” condition means

The object falls for some total time $T$ seconds. The phrase “during the last second” refers to the distance covered from time $(T-1)$ to time $T$.

Write distances using free-fall kinematics

For an object dropped from rest (no initial velocity), the distance fallen after time $t$ is $$s(t)=\tfrac12 gt^2.$$ So the total height is $$h=s(T)=\tfrac12 gT^2.$$ The distance traveled in the last second is $$s_{\text{last}}=s(T)-s(T-1)=\tfrac12 gT^2-\tfrac12 g(T-1)^2.$$ Simplify: $$s_{\text{last}}=\tfrac12 g\big(T^2-(T-1)^2\big) =\tfrac12 g\big(T^2-(T^2-2T+1)\big) =\tfrac12 g(2T-1) =g\left(T-\tfrac12\right).$$

Use the 36% condition to solve for $T$

We are told the last-second distance is $36\%$ of the total distance: $$g\left(T-\tfrac12\right)=0.36\left(\tfrac12 gT^2\right)=0.18gT^2.$$ Cancel $g$: $$T-\tfrac12=0.18T^2.$$ Rearrange: $$0.18T^2-T+0.5=0.$$ Multiply by 100 and simplify: $$18T^2-100T+50=0\quad\Rightarrow\quad 9T^2-50T+25=0.$$ Solve: $$T=\frac{50\pm\sqrt{50^2-4\cdot 9\cdot 25}}{2\cdot 9} =\frac{50\pm\sqrt{2500-900}}{18} =\frac{50\pm 40}{18}.$$ So $T=5$ s or $T=\tfrac{5}{9}$ s. We must have $T>1$ to have a “last second,” so $T=5$ s.

Find the height $h$

$$h=\tfrac12 gT^2=\tfrac12 g(25)=\tfrac{25}{2}g.$$ With $g=9.8\,\text{m/s}^2$: $$h=12.5\times 9.8=122.5\,\text{m}.$$