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Problem: Falling object from a given height

A red sphere at a certain height from the ground; vertical x-axis directed downward on which the position of the sphere (0) is shown; time (0) and velocity (0) are indicated. When the animation starts, the sphere starts falling, and it reaches the ground at instant 0.505 s when its position is 1.25 m and its velocity is 4.95 m/s.+0.00m/sx(m)0.00O0.000
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An object is left falling from a height of 1.25 m\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.

Solving the problem

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

h=1.25 mh = \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, gg.

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

Vertical x-axis with the positive direction going down.xO
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Let's consider the instant at which the object is left falling to be 0 and refer to it as t0t_0.

t0=0t_0 = 0

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

So, the position at the instant t0t_0 is

x0=0x_0 = 0
Vertical x-axis directed downward; the position x0 is at the origin.xO0x
1x

Since the object is initially at a height hh from the ground, the origin of the xx-axis is also at a height hh from the ground:

Vertical x-axis directed downward; the position x0 is at the origin; the distance from the origin to the ground below is h.xOhGround0x
1x

The object has no initial velocity, so

v0=0v_0 = 0

The object has a constant acceleration

a=g=9.81 m/s2a = g = \pu{9.81 m/s^2}

Since the object is moving with constant acceleration along a straight line, its velocity vv at an instant tt, subsequent to the instant t0t_0, is expressed as:

v=v0+a(tt0)v = v_0 + a(t - t_0)

Since t0=0t_0 = 0, v0=0v_0 = 0 and a=ga = g, the equation becomes:

v=0+g(t0)v = 0 + g(t - 0)

(1)v=gtv = gt \tag{1}

Again, since the object is moving with constant acceleration along a straight line, its position xx at an instant tt, subsequent to the instant t0t_0, is expressed as:

x=x0+v0(tt0)+12a(tt0)2x = x_0 + v_0(t - t_0) + \frac{1}{2} a(t - t_0)^2

And since t0=0t_0 = 0, x0=0x_0 = 0, v0=0v_0 = 0, and a=ga = g, the equation becomes:

x=0+0(t0)+12g(t0)2x = 0 + 0(t - 0) + \frac{1}{2} g(t - 0)^2

(2)x=12gt2x = \frac{1}{2} gt^2 \tag{2}

Finding the time it takes for the object to hit the ground

Eventually, at an instant t1t_1, the object hits the ground.

When that happens, the position of the object is

x1=hx_1 = h
Vertical x-axis directed downward; the position x0 is at the origin and the position x1 is below at the ground, distant h from the origin.xOhGround0x1x
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According to Eq. (2), the position at the instant t1t_1 when the object hits the ground is

x1=12gt12x_1 = \frac{1}{2} gt_1^2

Since x1=hx_1 = h, we can substitute x1x_1 with hh:

h=12gt12h = \frac{1}{2} gt_1^2

Let's now solve this equation for t1t_1:

2h=gt122h = gt_1^2

t12=2hgt_1^2 = \frac{2h}{g}

t1=±2hgt_1 = \pm \sqrt{\frac{2h}{g}}

We pick the positive solution because we know that t1>0t_1 \gt 0:

t1=2hgt_1 = \sqrt{\frac{2h}{g}}

Since the instant at which the object is left falling is 0, the instant t1t_1 coincides with the total time taken to hit the ground.

So, let's calculate it:

t1=2hgt_1 = \sqrt{\frac{2h}{g}}

t1=2(1.25 m)9.81 ms2t_1 = \sqrt{\frac{ 2(\pu{1.25 \cancel m}) }{ \pu{9.81 \cancel m//s^2} }}

t1=0.505 st_1 = \pu{0.505 s}

Finding the impact velocity of the object

Next, let's find the impact velocity v1v_1 that the object has at the instant t1t_1 when it hits the ground.

From Eq. (1), we know that the velocity at the instant t1t_1 is

v1=gt1v_1 = gt_1

And we know that the instant t1t_1 is

t1=2hgt_1 = \sqrt{\frac{2h}{g}}

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

v1=g2hgv_1 = g \sqrt{\frac{2h}{g}}

v1=2hgg2v_1 = \sqrt{\frac{2h}{\cancel g} g^{\cancel 2}}

v1=2hgv_1 = \sqrt{2hg}

v1=2(1.25 m)(9.81 ms2)v_1 = \sqrt{ 2(\pu{1.25 m}) (\pu{9.81 m//s^2}) }

v1=4.95 msv_1 = \pu{4.95 m//s}

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

Drawing the acceleration vs time graph

Since the object has a constant acceleration

a=g=9.81 m/s2a = g = \pu{9.81 m/s^2}

The acceleration vs time graph looks like this:

Acceleration vs time graph consisting of a horizontal line.0.10.20.30.40.50ts()asm/)(22468100
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Drawing the velocity vs time graph

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

v=gtv = gt

Since vv is a linear function of tt, 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 t0=0t_0 = 0, the object has a velocity v0=0v_0 = 0, and at the instant t1=0.505 st_1 = \pu{0.505 s} when the object hits the ground, the velocity is v1=4.95 m/sv_1 = \pu{4.95 m/s}.

We can now represent the velocity vs time graph:

Velocity vs time graph consisting of an ascending line that passes through the origin.0.10.20.30.40.50ts()123450v(sm/)
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Drawing the position vs time graph

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

x=12gt2x = \frac{1}{2} gt^2

Since xx is a quadratic function of tt, the position vs time graph is a parabola.

The coefficient of t2t^2 is positive, so the parabola opens upward.

Since the quadratic function has only a t2t^2, the vertex of the parabola is at t=0t = 0, and at that instant, the position is x=0x = 0.

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

At the instant t1=0.505 st_1 = \pu{0.505 s} when the object hits the ground, the position is x1=1.25 mx_1 = \pu{1.25 m}.

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

x(0.1 s)=12(9.81 ms2)(0.1 s)2x_{(\pu{0.1 s})} = \frac{1}{2} (\pu{9.81 m//s^2}) (\pu{0.1 s})^2

x(0.1 s)=0.0491 mx_{(\pu{0.1 s})} = \pu{0.0491 m}

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

x(0.2 s)=12(9.81 ms2)(0.2 s)2x_{(\pu{0.2 s})} = \frac{1}{2} (\pu{9.81 m//s^2}) (\pu{0.2 s})^2

x(0.2 s)=0.196 mx_{(\pu{0.2 s})} = \pu{0.196 m}

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

x(0.4 s)=12(9.81 ms2)(0.4 s)2x_{(\pu{0.4 s})} = \frac{1}{2} (\pu{9.81 m//s^2}) (\pu{0.4 s})^2

x(0.4 s)=0.785 mx_{(\pu{0.4 s})} = \pu{0.785 m}

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

Position vs time graph consisting of the right half of a parabola that opens upward with the vertex in the origin.0.10.20.30.40.50ts()0.250.500.751.001.250x(m)
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