Question

An object is dropped from a height h. During the last second of its journey the object travels a
distance 9h/25.Then,
What is the value of ‘h’?

Answer 1

An object is dropped from a height $h$. During the last second of its journey the object travels a distance $9h/25$. Then, what is the value of 'h'?

Solution:

The equation of motion for the object on the path of all way:

$$h = \frac{g t^2}{2} \Rightarrow t = \sqrt{\frac{2h}{g}} (1) - \text{time of the journey}$$

The equation of motion for an object that flies time $t - 1$ (after one second object will end movement):

$$h - h_1 = \frac{g(t - 1)^2}{2}; h_1 = \frac{9}{25} h$$ $$h - \frac{9}{25} h = \frac{g(t - 1)^2}{2}$$ $$\frac{16}{25} h = \frac{g(t - 1)^2}{2}$$ $$\sqrt{\frac{16}{25}} h = \sqrt{\frac{g(t - 1)^2}{2}}$$ $$\frac{4}{5} \sqrt{h} = \frac{\sqrt{g}(t - 1)}{\sqrt{2}} (2)$$ $$(1) \text{in } (2): \frac{4}{5} \sqrt{h} = \frac{\sqrt{g} \left( \sqrt{\frac{2h}{g}} - 1 \right)}{\sqrt{2}}$$ $$4 \sqrt{2} \sqrt{h} = 5 (\sqrt{2h} - \sqrt{g})$$ $$\sqrt{h} (5 \sqrt{2} - 4 \sqrt{2}) = 5 \sqrt{g}$$ $$\sqrt{h} = \frac{5 \sqrt{g}}{\sqrt{2}} \Rightarrow h = \frac{5g}{2} = \frac{5 \cdot 9.8 \frac{m}{sec^2}}{2} = 24.5m$$

Answer: $h = 24.5m$