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An object is left falling from a height of $\pu{1.25 m}$.

- Find the time the falling object takes to hit the ground.
- Find the velocity that the object has when it hits the ground (i.e., the impact velocity).
- Draw the acceleration vs time graph, the velocity vs time graph, and the position vs time graph that describe the motion of the object from the moment it is left falling to the moment it hits the ground.

We have an object that is left falling from a given height

$h = \pu{1.25 m}$

The object has no initial velocity.

The object moves vertically downward with constant acceleration until it hits the ground.

The constant acceleration is the gravitational acceleration, $g$.

Let's start by representing the $x$-axis, along which the object moves, with the positive direction coinciding with the direction of motion of the object, i.e., vertically downward:

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Let's consider the instant at which the object is left falling to be 0 and refer to it as $t_0$.

$t_0 = 0$

And let's consider the origin of the $x$-axis to be right where the object is the moment it is released.

So, the position at the instant $t_0$ is

$x_0 = 0$

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Since the object is initially at a height $h$ from the ground, the origin of the $x$-axis is also at a height $h$ from the ground:

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The object has no initial velocity, so

$v_0 = 0$

The object has a constant acceleration

$a = g = \pu{9.81 m/s^2}$

Since the object is moving with constant acceleration along a straight line, its velocity $v$ at an instant $t$, subsequent to the instant $t_0$, is expressed as:

$v = v_0 + a(t - t_0)$

Since $t_0 = 0$, $v_0 = 0$ and $a = g$, the equation becomes:

$v = 0 + g(t - 0)$

$v = gt \tag{1}$

Again, since the object is moving with constant acceleration along a straight line, its position $x$ at an instant $t$, subsequent to the instant $t_0$, is expressed as:

$x = x_0 + v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2$

And since $t_0 = 0$, $x_0 = 0$, $v_0 = 0$, and $a = g$, the equation becomes:

$x = 0 + 0(t - 0) + \frac{1}{2} g(t - 0)^2$

$x = \frac{1}{2} gt^2 \tag{2}$

Eventually, at an instant $t_1$, the object hits the ground.

When that happens, the position of the object is

$x_1 = h$

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According to Eq. (2), the position at the instant $t_1$ when the object hits the ground is

$x_1 = \frac{1}{2} gt_1^2$

Since $x_1 = h$, we can substitute $x_1$ with $h$:

$h = \frac{1}{2} gt_1^2$

Let's now solve this equation for $t_1$:

$2h = gt_1^2$

$t_1^2 = \frac{2h}{g}$

$t_1 = \pm \sqrt{\frac{2h}{g}}$

We pick the positive solution because we know that $t_1 \gt 0$:

$t_1 = \sqrt{\frac{2h}{g}}$

Since the instant at which the object is left falling is 0, the instant $t_1$ coincides with the total time taken to hit the ground.

So, let's calculate it:

$t_1 = \sqrt{\frac{2h}{g}}$

$t_1 = \sqrt{\frac{ 2(\pu{1.25 \cancel m}) }{ \pu{9.81 \cancel m//s^2} }}$

$t_1 = \pu{0.505 s}$

Next, let's find the impact velocity $v_1$ that the object has at the instant $t_1$ when it hits the ground.

From Eq. (1), we know that the velocity at the instant $t_1$ is

$v_1 = gt_1$

And we know that the instant $t_1$ is

$t_1 = \sqrt{\frac{2h}{g}}$

If we substitute this in the previous equation, we find that

$v_1 = g \sqrt{\frac{2h}{g}}$

$v_1 = \sqrt{\frac{2h}{\cancel g} g^{\cancel 2}}$

$v_1 = \sqrt{2hg}$

$v_1 = \sqrt{ 2(\pu{1.25 m}) (\pu{9.81 m//s^2}) }$

$v_1 = \pu{4.95 m//s}$

Thus, we found that the object takes $\pu{0.505 s}$ to hit the ground and the impact velocity that it has is $\pu{4.95 m/s}$.

Since the object has a constant acceleration

$a = g = \pu{9.81 m/s^2}$

The acceleration vs time graph looks like this:

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The velocity of the object at an instant $t$ is given by Eq. (1):

$v = gt$

Since $v$ is a linear function of $t$, the velocity vs time graph is a straight line.

Thus, we only need to know two points to represent the graph.

We already know them: at the instant $t_0 = 0$, the object has a velocity $v_0 = 0$, and at the instant $t_1 = \pu{0.505 s}$ when the object hits the ground, the velocity is $v_1 = \pu{4.95 m/s}$.

We can now represent the velocity vs time graph:

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The position of the object at an instant $t$ is given by Eq. (2):

$x = \frac{1}{2} gt^2$

Since $x$ is a quadratic function of $t$, the position vs time graph is a parabola.

The coefficient of $t^2$ is positive, so the parabola opens upward.

Since the quadratic function has only a $t^2$, the vertex of the parabola is at $t = 0$, and at that instant, the position is $x = 0$.

Let's determine a few points of the parabola and then draw it.

At the instant $t_1 = \pu{0.505 s}$ when the object hits the ground, the position is $x_1 = \pu{1.25 m}$.

At the instant $t = \pu{0.1 s}$, the position is

$x_{(\pu{0.1 s})} = \frac{1}{2} (\pu{9.81 m//s^2}) (\pu{0.1 s})^2$

$x_{(\pu{0.1 s})} = \pu{0.0491 m}$

At the instant $t = \pu{0.2 s}$, the position is

$x_{(\pu{0.2 s})} = \frac{1}{2} (\pu{9.81 m//s^2}) (\pu{0.2 s})^2$

$x_{(\pu{0.2 s})} = \pu{0.196 m}$

At the instant $t = \pu{0.4 s}$, the position is

$x_{(\pu{0.4 s})} = \frac{1}{2} (\pu{9.81 m//s^2}) (\pu{0.4 s})^2$

$x_{(\pu{0.4 s})} = \pu{0.785 m}$

We now have enough points to draw the position vs time graph:

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