A stone dropped from the top of the tower touches the ground in $4 ext{ sec}$. The height of the tower is ab

Question

A stone dropped from the top of the tower touches the ground in $4 	ext{ sec}$. The height of the tower is about:

Accepted Answer

1.  **Analyze Initial Conditions:** The stone is dropped, implying an initial velocity $u = 0$. The time taken to reach the ground is $t = 4 	ext{ s}$. Acceleration due to gravity $a = g$. To estimate 'about' the height, we can approximate $g \approx 10 	ext{ m/s}^2$ (or use $9.8 	ext{ m/s}^2$).
2.  **Apply Kinematic Equation:** Use the equation for displacement in uniformly accelerated motion: $h = ut + \frac{1}{2}gt^2$ .
3.  **Calculation:**
    $h = (0)(4) + \frac{1}{2}(10)(4)^2$
    $h = 0 + 5(16) = 80 	ext{ m}$.
    (Note: Using precise $g=9.8 	ext{ m/s}^2$, $h = 0.5(9.8)(16) = 78.4 	ext{ m}$, which rounds to $80 	ext{ m}$).

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