An object falls freely from height $h$ above the ground. It travels $\frac{5}{9}h$ of the total height in the

Question

An object falls freely from height $h$ above the ground. It travels $\frac{5}{9}h$ of the total height in the last $1	ext{ s}$. The height $h$ is: (use $g=10	ext{ m/s}^2$)

Accepted Answer

1.  **Define Variables:** Let $t$ be the total time taken to fall through height $h$. Let $u=0$ (starts from rest) and $a=g$.
2.  **Total Distance Equation:** Using the second equation of motion :
    $$h = \frac{1}{2}gt^2$$
3.  **Distance in (t-1) seconds:** The object travels $\frac{5}{9}h$ in the last second. Therefore, the distance traveled in the first $(t-1)$ seconds is:
    $$h' = h - \frac{5}{9}h = \frac{4}{9}h$$
    Using the kinematic equation for this duration:
    $$\frac{4}{9}h = \frac{1}{2}g(t-1)^2$$
4.  **Solve for Time (t):** Substitute $h = \frac{1}{2}gt^2$ into the second equation:
    $$\frac{4}{9}\left(\frac{1}{2}gt^2ight) = \frac{1}{2}g(t-1)^2$$
    Divide both sides by $\frac{1}{2}g$:
    $$\frac{4}{9}t^2 = (t-1)^2$$
    Take the square root of both sides:
    $$\frac{2}{3}t = t - 1 \quad 	ext{or} \quad \frac{2}{3}t = -(t-1)$$
    Solving $\frac{2}{3}t = t - 1$ gives $1 = t - \frac{2}{3}t \Rightarrow 1 = \frac{1}{3}t \Rightarrow t = 3	ext{ s}$. (The other solution yields $t < 1$, which is invalid for the 'last 1 second' condition).
5.  **Calculate Height (h):** Substitute $t=3	ext{ s}$ back into the height equation:
    $$h = \frac{1}{2}(10)(3)^2 = 5 	imes 9 = 45	ext{ m}$$

Step-by-Step Solution

Practice All PHYSICS Questions