Free fall is the motion of an object under the influence of gravitational force only. The object accelerates downward at a constant rate, regardless of its mass.
Key Equations for Free Fall:
\[v = v_0 + gt\]
\[y = y_0 + v_0t + \frac{1}{2}gt^2\]
\[v^2 = v_0^2 + 2g(y - y_0)\]
Where:
Note: We typically take upward as positive direction, so \(g = -9.8 \text{ m/s}^2\). However, for simplicity in magnitude calculations, we often use \(g = 9.8 \text{ m/s}^2\) downward.
All solutions assume \(g = 9.8 \text{ m/s}^2\) and negligible air resistance unless otherwise specified.
Problems progress from basic to advanced levels to build understanding of free fall kinematics.
A stone is dropped from the top of a building that is 80 meters high. Assume no air resistance and \(g = 9.8 \text{ m/s}^2\).
Given: \(y_0 = 80 \text{ m}\), \(v_0 = 0 \text{ m/s}\), \(g = 9.8 \text{ m/s}^2\), \(y = 0 \text{ m}\) (ground level)
(a) Time to reach the ground:
Step 1: Use the position equation with \(y = 0\) (ground level):\[y = y_0 + v_0t + \frac{1}{2}gt^2\]
\[0 = 80 + 0 \cdot t + \frac{1}{2}(-9.8)t^2\]
\[4.9t^2 = 80\]
\[t^2 = \frac{80}{4.9} = 16.3265\]
\[t = \sqrt{16.3265} = 4.04 \text{ s}\]
Step 2: Verify with the time equation:\[v = v_0 + gt = 0 + (-9.8)(4.04) = -39.6 \text{ m/s}\]
(b) Velocity before impact:
Step 1: Use the velocity equation:\[v^2 = v_0^2 + 2g(y - y_0)\]
\[v^2 = 0 + 2(-9.8)(0 - 80)\]
\[v^2 = 1568\]
\[v = -\sqrt{1568} = -39.6 \text{ m/s}\]
(Negative sign indicates downward direction)
(a) The stone takes \(\mathbf{4.04 \text{ seconds}}\) to reach the ground.
(b) The stone's velocity just before impact is \(\mathbf{39.6 \text{ m/s}}\) downward.
A ball is thrown straight downward from a height of 45 meters with an initial velocity of 5 m/s.
Given: \(y_0 = 45 \text{ m}\), \(v_0 = -5 \text{ m/s}\) (downward), \(g = 9.8 \text{ m/s}^2\), \(y = 0 \text{ m}\)
(a) Time to hit the ground:
\[y = y_0 + v_0t + \frac{1}{2}gt^2\]
\[0 = 45 + (-5)t + \frac{1}{2}(-9.8)t^2\]
\[0 = 45 - 5t - 4.9t^2\]
\[4.9t^2 + 5t - 45 = 0\]
\[t = \frac{-5 \pm \sqrt{5^2 - 4(4.9)(-45)}}{2(4.9)}\]
\[t = \frac{-5 \pm \sqrt{25 + 882}}{9.8}\]
\[t = \frac{-5 \pm \sqrt{907}}{9.8}\]
\[t = \frac{-5 \pm 30.12}{9.8}\]
We take the positive root: \(t = \frac{-5 + 30.12}{9.8} = \frac{25.12}{9.8} = 2.56 \text{ s}\)
(b) Impact velocity:
\[v = v_0 + gt\]
\[v = -5 + (-9.8)(2.56)\]
\[v = -5 - 25.09 = -30.09 \text{ m/s}\]
(a) The ball takes \(\mathbf{2.56 \text{ seconds}}\) to hit the ground.
(b) The impact velocity is \(\mathbf{30.1 \text{ m/s}}\) downward.
A ball is thrown vertically upward from ground level with an initial velocity of 25 m/s.
Given: \(y_0 = 0 \text{ m}\), \(v_0 = 25 \text{ m/s}\) (upward), \(g = 9.8 \text{ m/s}^2\) downward
(a) Maximum height:
\[v^2 = v_0^2 + 2g(y - y_0)\]
\[0 = (25)^2 + 2(-9.8)(y - 0)\]
\[0 = 625 - 19.6y\]
\[19.6y = 625\]
\[y = \frac{625}{19.6} = 31.89 \text{ m}\]
(b) Time to reach maximum height:
\[0 = 25 + (-9.8)t\]
\[9.8t = 25\]
\[t = \frac{25}{9.8} = 2.55 \text{ s}\]
(c) Total time in air:
\[0 = 0 + 25t + \frac{1}{2}(-9.8)t^2\]
\[0 = 25t - 4.9t^2\]
\[0 = t(25 - 4.9t)\]
\(t = 0\) (initial time) or \(t = \frac{25}{4.9} = 5.10 \text{ s}\)
(a) Maximum height: \(\mathbf{31.9 \text{ meters}}\)
(b) Time to reach maximum height: \(\mathbf{2.55 \text{ seconds}}\)
(c) Total time in air: \(\mathbf{5.10 \text{ seconds}}\)
A rock is thrown vertically upward from the edge of a cliff that is 75 meters above ground level. The initial velocity is 15 m/s upward.
Given: \(y_0 = 75 \text{ m}\), \(v_0 = 15 \text{ m/s}\) (upward), \(g = 9.8 \text{ m/s}^2\), \(y = 0 \text{ m}\) (ground)
(a) Maximum height above ground:
\[v^2 = v_0^2 + 2g(y - y_0)\]
\[0 = (15)^2 + 2(-9.8)(y - 75)\]
\[0 = 225 - 19.6(y - 75)\]
\[19.6(y - 75) = 225\]
\[y - 75 = \frac{225}{19.6} = 11.48\]
\[y = 75 + 11.48 = 86.48 \text{ m}\]
(b) Time to hit the ground:
\[0 = 75 + 15t + \frac{1}{2}(-9.8)t^2\]
\[0 = 75 + 15t - 4.9t^2\]
\[4.9t^2 - 15t - 75 = 0\]
\[t = \frac{15 \pm \sqrt{(-15)^2 - 4(4.9)(-75)}}{2(4.9)}\]
\[t = \frac{15 \pm \sqrt{225 + 1470}}{9.8}\]
\[t = \frac{15 \pm \sqrt{1695}}{9.8}\]
\[t = \frac{15 \pm 41.17}{9.8}\]
We take the positive root: \(t = \frac{15 + 41.17}{9.8} = \frac{56.17}{9.8} = 5.73 \text{ s}\)
(c) Velocity before impact:
\[v = v_0 + gt\]
\[v = 15 + (-9.8)(5.73)\]
\[v = 15 - 56.15 = -41.15 \text{ m/s}\]
(a) Maximum height above ground: \(\mathbf{86.5 \text{ meters}}\)
(b) Time to hit the ground: \(\mathbf{5.73 \text{ seconds}}\)
(c) Velocity before impact: \(\mathbf{41.2 \text{ m/s}}\) downward
A stone is dropped from the top of a 100-meter tall building. One second later, another stone is thrown vertically downward from the same point with an initial velocity of 20 m/s.
Given: Building height = 100 m, \(g = 9.8 \text{ m/s}^2\)
Let \(t\) = time after first stone is dropped.
(a) Time when both stones are at same height:
\[y_1 = 100 - \frac{1}{2}gt^2\]
\[y_2 = 100 - 20(t-1) - \frac{1}{2}g(t-1)^2 \quad \text{for} \quad t \geq 1\]
Step 3: Set \(y_1 = y_2\):\[100 - \frac{1}{2}gt^2 = 100 - 20(t-1) - \frac{1}{2}g(t-1)^2\]
\[-\frac{1}{2}gt^2 = -20(t-1) - \frac{1}{2}g(t-1)^2\]
\[\frac{1}{2}g[t^2 - (t-1)^2] = 20(t-1)\]
\[\frac{1}{2}g[(t^2 - (t^2 - 2t + 1)] = 20(t-1)\]
\[\frac{1}{2}g(2t - 1) = 20(t-1)\]
\[g(2t - 1) = 40(t-1)\]
\[9.8(2t - 1) = 40t - 40\]
\[19.6t - 9.8 = 40t - 40\]
\[40t - 19.6t = 40 - 9.8\]
\[20.4t = 30.2\]
\[t = \frac{30.2}{20.4} = 1.48 \text{ s}\]
(b) Height at that time:
Step 1: Use \(y_1\) equation:\[y = 100 - \frac{1}{2}(9.8)(1.48)^2\]
\[y = 100 - 4.9(2.1904)\]
\[y = 100 - 10.73 = 89.27 \text{ m}\]
(c) Which hits first and time difference:
Step 1: Time for first stone to hit ground:\[0 = 100 - \frac{1}{2}gt_1^2 \Rightarrow t_1 = \sqrt{\frac{200}{9.8}} = \sqrt{20.41} = 4.52 \text{ s}\]
Step 2: Time for second stone to hit ground:\[0 = 100 - 20t_2 - \frac{1}{2}gt_2^2\]
\[4.9t_2^2 + 20t_2 - 100 = 0\]
\[t_2 = \frac{-20 \pm \sqrt{400 + 1960}}{9.8} = \frac{-20 \pm \sqrt{2360}}{9.8}\]
\[t_2 = \frac{-20 \pm 48.58}{9.8}\]
Positive root: \(t_2 = \frac{28.58}{9.8} = 2.92 \text{ s}\)
But this is time after second stone is thrown. Total time after first stone dropped: \(t_2' = 1 + 2.92 = 3.92 \text{ s}\)
Step 3: Time difference:\[ \Delta t = 4.52 - 3.92 = 0.60 \text{ s}\]
(a) Both stones are at the same height \(\mathbf{1.48 \text{ seconds}}\) after the first stone is dropped.
(b) Height above ground at that time: \(\mathbf{89.3 \text{ meters}}\)
(c) The second stone hits first, \(\mathbf{0.60 \text{ seconds}}\) before the first stone.