Lesson Explainer: Vertical Motion under Gravity Mathematics

In this explainer, we will learn how to use the kinematics equations of uniform acceleration to model the vertical motion of a body with uniform acceleration due to gravity.

Near the surface of Earth, a body that is not acted on by a net vertical force other than its own weight will accelerate uniformly vertically downward. The acceleration due to gravity is represented by 𝑔 and has a magnitude of approximately 9.8 m/s2. Acceleration due to gravity varies with the distance from the center of mass of Earth, but the variation in 𝑔 with altitude is small near the surface of Earth and is treated as a constant value.

It is important to distinguish between acceleration due to gravity near the surface of Earth and the net acceleration of a body near the surface of Earth. A body near the surface of Earth does not necessarily accelerate vertically downward at 9.8 m/s2. For example, a body at rest, or moving parallel to the surface of Earth (modeling Earth as a sphere), has zero vertical acceleration. This is because the weight of a body in contact with the surface of Earth is not the only force acting on the body.

This explainer exclusively considers bodies on which the only force acting is the force of their weights, so we will neglect resistive forces exerted on bodies by air. The motion of such bodies can be modeled using kinematic equations for motion with uniform acceleration, such as 𝑣=𝑢+𝑎𝑡,𝑣=𝑢+2𝑎𝑠,𝑠=𝑢𝑡+12𝑎𝑡, where 𝑢 is the initial velocity of the particle, 𝑣 is the final velocity of the particle, 𝑎 is the acceleration of the particle, and 𝑠 is the displacement of the particle.

Let us look at an example of a body moving freely that is acted on only by its weight.

Example 1: Finding the Initial Velocity of a Particle That Is Projected Vertically Upward

A particle was projected vertically upward from the ground. Given that the maximum height the particle reached was 62.5 m, find the velocity at which it was projected. Take the acceleration due to gravity 𝑔=9.8/ms.

Answer

The motion of the particle can be modeled using the equation 𝑣=𝑢+2𝑎𝑠.

The acceleration of the particle is only due to gravity, so the magnitude of the acceleration is 9.8 m/s2. The acceleration is vertically downward. The particle is projected vertically upward, so acceleration due to gravity is in the opposite direction to the initial velocity of the particle. If the direction of the initial velocity of the particle is taken as positive, the acceleration is negative; hence, we have that 𝑣=𝑢2(9.8)(𝑠)=𝑢19.6(𝑠).

This equation can be rearranged to make 𝑢 the subject: 𝑢=𝑣+19.6(𝑠).

The particle is instantaneously at rest when it reaches its maximum height, so at that instant, it has zero velocity. We see then that 𝑢=0+19.6(62.5)=1225.

Taking the positive root of 𝑢, we obtain 𝑢=1225=35/.ms

The negative root is neglected as this corresponds to vertically downward velocity of the particle, and the particle cannot have initially vertically downward velocity as it is projected vertically upward.

A body that is uniformly accelerated has a continuously varying velocity, but the average value of this velocity is the mean of its velocities before and after acceleration, as given by 𝑣=𝑣+𝑢2.average

Let us look at an example where the average velocity of a body accelerated only by gravity is determined.

Example 2: Finding the Average Velocity of a Falling Body

If a body, which was dropped from a building, took 3 seconds to reach the ground, find its average velocity as it fell. Let the acceleration due to gravity 𝑔=9.8/ms.

Answer

The acceleration of a body accelerating uniformly is given by 𝑎=𝑣𝑢𝑡.

This expression can be rearranged to determine the average velocity as follows: 𝑣𝑢=𝑎𝑡𝑣=𝑢+𝑎𝑡.

Substituting the known values of 𝑡 and 𝑎, and assuming that 𝑢 is zero as the body was dropped rather than projected vertically, we have that 𝑣=9.8(3)=29.4/.ms

The average velocity of the object is the mean of its velocities before and after acceleration, given by 𝑣=𝑣+𝑢2.average

We have already seen that 𝑢 is zero, so we have that 𝑣=29.42=14.7/.averagems

Now, let us look at an example where we determine the displacement of a body projected vertically with a known velocity.

Example 3: Finding the Maximum Height Reached by a Ball Projected Vertically Upward

A particle is projected vertically upward at 7 m/s from a point 38.7 m above the ground. Find the maximum height the particle can reach. Consider the acceleration due to gravity to be 𝑔=9.8/ms.

Answer

The particle reaches its maximum height at the instant that it is instantaneously at rest and about to start falling back toward the point from which it was projected.

In this example, we are given the initial velocity of the particle, the initial vertical displacement of the particle, and the vertical acceleration of the particle. The velocity of the particle at its greatest vertical displacement will be instantaneously zero.

The displacement of the particle from the point it was projected at this instant can be determined using a formula containing these terms; hence, we can use the formula 𝑣=𝑢+2𝑎𝑠, where 𝑣=0/ms, 𝑢=7/ms, and 𝑎=9.8/ms. Rearranging the equation to make 𝑠 the subject, we obtain 𝑠=𝑣𝑢2𝑎.

Substituting known values, we find 𝑠=072(9.8)=2.5.m

The initial vertical displacement of the particle was 38.7 m, so the greatest vertical displacement of the particle is given by 38.7+2.5=41.2.m

Let us look at another such example.

Example 4: Using the Equations of Motion to Solve a Vertical Projection Question

If a body is projected vertically upward with speed 𝑉 to reach maximum height , then the speed the body should be projected at to reach height 4 is .

Answer

The particle reaches its maximum height at the instant that it is instantaneously at rest and about to start falling back toward the point from which it was projected. The displacement of the particle from the point it was projected at this instant can be determined using the formula 𝑣=𝑢+2𝑎𝑠.

Rearranging the equation to make 𝑠 the subject and noting both that 𝑣 is zero and that because the vertically downward direction is taken as negative, 𝑢 and 𝑎 are negative, we therefore obtain 𝑠=𝑢2𝑔.

The numerator and denominator of the right side of the expression are negative, so it is equivalent to the expression 𝑠=𝑢2𝑔.

For a value of 𝑢 equal to 𝑉, the value of 𝑠 is equal to . It is important to note that we must not confuse the value of the initial velocity 𝑉 with the symbol used for the final velocity term, 𝑣.

Using the values of 𝑉 and , we have that =𝑉2𝑔.

We need to determine how an initial velocity, denoted by 𝑉, corresponding to a value for 𝑠 of 4 compares to the initial velocity 𝑉.

We know that a particle launched vertically upward with a velocity 𝑉 has a velocity of zero when its vertical displacement is 4.

We can use the equation 𝑣=𝑢+2𝑎𝑠 and substitute the values of 𝑉 for initial velocity, 𝑔 for acceleration, and 4 for displacement. Noting that 𝑣 is zero and that 𝑔 is negative when 𝑉 is positive, we have that 0=𝑉2𝑔(4).

Substituting =𝑉2𝑔, we have that 0=𝑉2𝑔(4)𝑉2𝑔0=𝑉8𝑔𝑉2𝑔0=𝑉4𝑉.

We can rearrange this expression to obtain an expression that relates 𝑉 and 𝑉: 4𝑉=𝑉.

Taking square roots of both sides, we find that 4𝑉=𝑉2𝑉=𝑉.

We see that 𝑉 has twice the value of 𝑉.

Let us look at an example where a body is projected vertically with an unknown velocity and the final velocity of the body is also unknown.

Example 5: Finding the Time It Takes for a Body to Reach the Base of a Tower

A body was projected vertically downward from the top of a tower whose height is 80 m. Given that it covered 35.9 m during the 1st second of its motion, find the time taken to reach the ground rounded to the nearest two decimal places. Let the acceleration due to gravity 𝑔=9.8/ms.

Answer

Neither the initial velocity of the body nor its velocity at any point during its motion can be determined from the information given in the question, and so the answer cannot be determined by comparing the initial and final velocities of the body.

It might seem that the final velocity of the body is zero, as the body eventually reaches the ground, but at the instant that the body reaches the ground, its instantaneous velocity is not zero. If the body does not rebound, the velocity of the body a short time after it reaches the ground will be zero, but it does not reach the ground with zero velocity.

The velocities of the body are not known, but the initial velocity of the body can be determined using the formula 𝑠=𝑢𝑡+12𝑎𝑡 and rearranging to make 𝑢 the subject to get 𝑢=𝑠𝑎𝑡𝑡.

The initial velocity is in the same direction as the acceleration. We take this as the positive direction.

Substituting known values, we find that 𝑢=35.99.8(1)1=35.94.9=31/.ms

Knowing the initial velocity, the final velocity can be determined using the formula 𝑣=𝑢+2𝑎𝑠 as follows: 𝑣=31+2(9.8)80=2529𝑣=2529/.ms

We can now rearrange the equation 𝑣=𝑢+𝑎𝑡 to make 𝑡 the subject, obtaining 𝑡=𝑣𝑢𝑎𝑡=2529319.8.

To two decimal places, 𝑡 is 1.97 seconds.

Let us summarize what we have learned from these examples.

Key Points

  • Near the surface of Earth, a body that is not acted on by a net vertical force other than its own weight will uniformly accelerate vertically downward at 9.8 m/s2.
  • The motion of a body accelerated only by gravity can be modeled using kinematic equations of motion with constant acceleration, where the acceleration has a magnitude of 9.8 m/s2.
  • A body moving vertically that is accelerated only by gravity can have an initial velocity either in the same direction as or in the opposite direction to its acceleration.

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