### A ball is thrown vertically upward – FAQ

##### April 7, 2017

## A ball is thrown vertically upward –Discussion on different phases

A ball is thrown vertically upward: You must have seen someone **throwing a ball vertically upwards** or you have thrown it yourself.

Yes, when a ball is thrown vertically upwards it goes up for a while and then stops and starts to come down and touches the ground with a bang if you can’t catch it.

Let’s discuss the phases of this traversal and motion with some formula and examples.

## object thrown vertically upward formula – quick reference

If a ball is thrown vertically upwards with an initial velocity V_{0} then here is a set of formula for your quick reference.

1) Maximum height reached =

H = V_{0}^{2} / (2 g)

2) Velocity at the highest point = 0

3) Time for upward movement = V_{0} /g

4) Time for downward movement =

V_{0} /g

5) Total time of travel in air = (2 V_{0} )/g

## A ball is thrown vertically upwards – under earth’s gravity

Now its time to discuss in details and derive the above equations. Please note that if you derive the equations you will get a better understanding of the concept.

Please note that this motion of the ball is entirely due to the **earth’s gravity** on the ball.

Here we will discuss about the acceleration of the ball thrown vertically upwards – both during its upward and downward motion.

This acceleration is obviously the **acceleration due to gravity** (denoted by g) caused by the earth’s gravity**.**

Note that: Here we are not considering the air resistance.

Anyways, in our discussion we will also consider the **maximum height reached** and the **time taken for upward and downward movements**.

Fair enough? So let’s start.

**Upward movement of the ball **when a ball is thrown vertically upward

Say a ball is thrown vertically upward with some velocity say **v1**, which we will consider as the initial velocity for the upward path.

After certain time period** **say** t,** the ball reaches a height beyond which it can’t move upwards anymore and stops there i.e.** its velocity becomes zero at that height.**

The height where the velocity becomes zero which is the **maximum height** the ball went upward, say is **H**.

And for this upward movement, the **final velocity v2 is 0** because the ball has stopped at the end of this upward traversal.

### Why an object thrown upwards comes down after reaching a point?

When an object is thrown with certain initial velocity (say V), it gains a Kinetic energy at that moment of throwing.

As it moves upwards from its initial position (where from it’s thrown) and gains height, its potential energy rises.

This happens because Potential Energy (PE) is directly proportional to the height of the object. (PE = mgh where h is the height).

Now from the Law of conversation of energy, we can say this rise in PE is happening at the cost of some form of energy being transformed.

Here it’s kinetic energy of the object which is expressed as 0.5 m V^2.

As height rises, velocity falls which results in reduction of KE and corresponding rise in PE.

At one point KE becomes zero. At that point velocity becomes zero.

This is called the highest point for an upward vertical movement. After this the ball starts falling downwards.

Differently we can say that the KE availed by the thrown object gets corroded under the negative influence of oppositely directing gravity (Gravitational force due to earth).

The influence is negative because gravity is pulling downwards while the ball is trying to move upwards.

After some time when the entire KE gets nullified, the ball stops. And then starts falling towards the earth’s surface.

In all the above discussions, we have considered Air Resistance as negligible.

**Those who are aware of escape velocity, you can read a post on it here: **Escape Velocity**

### Forces acting on a ball thrown upwards

Considering the Air resistance or Drag force negligible, the only force acting on the ball is Gravity i.e. the **Gravitational Pull** of the earth towards the the centre of the it.

### Acceleration of a ball thrown vertically upwards-during upward movement

It’s pretty evident that after the upward throw, the velocity of the ball gradually decreases i.e. a negative acceleration was working on the ball.

The acceleration is negative because this acceleration is directing downwards while the ball is moving upward.

And because of this negative acceleration the velocity of the ball is gradually decreasing.

And Yes. this acceleration is nothing but the **acceleration due to gravity** caused by gravitational pull or force exerted by the earth on the ball.

It’s value is generally taken as 9.8 m/s^2.

See here **how acceleration due to gravity varies with height and depth **wrt the surface of the earth.

### time taken by the upward movement

As this **acceleration due to gravity (g)** is working opposite to the upward velocity we have to use a negative sign in the formula below, used for the upward movement of the ball.

We know the value of g in SI is 9.8 m/second square.

Using one of the equations of motion,

**v2 = v1 – gt ……………………(i)**

**As v2=0**, (at the highest point the velocity becomes zero), then we can write the previous equation as follows:

** 0 = v1 – gt**

or, **t = v1/g **…………………….(ii)

So from equation (ii), the **time taken by the ball to reach the maximum height** is expressed as

= (Initial Velocity with which the ball is thrown vertically upwards) / (acceleration due to gravity)…..(iii)

So just for example, if a ball is thrown vertically upwards with 98 m/s velocity, then to reach the maximum height it will take = 98/9.8 =10 seconds.

### maximum height reached during upward movement when a ball is thrown vertically upward

And the maximum height H reached is obtained from the formula:

**v2 ^{2}=v1^{2}-2gH ( under negative acceleration) ……………… (iii)**

As v2=0, (**at the highest point the velocity becomes zero**), so we can rewrite equation iii as:

0 = v1^{2 }– 2gH

**or H= v1 ^{2}/2g (equation of maximum height) ………. (iv)**

Therefore if a ball is thrown vertically upwards with 98 m/s velocity, the maximum height reached by it would be = (98 x98 )/(2 x 9.8) meter = 490 meter.

#### What are the velocity and acceleration of the ball when it reaches the highest point?

Just summing up the answer here though we have already gone through this in the section above.

The **velocity at the highest point** is zero as the ball momentarily halts there before starting its downward movement.

And the **acceleration working on the ball at this point** is the acceleration due to gravity (g) and this time it’s considered positive i.e. downwards as the ball is now ready to free fall. (ignoring air resistance)

**Downward movement of the ball**

### acceleration of a ball thrown vertically upwards-during downward movement

As the ball reaches the maximum height now it starts its** free fall **towards the earth.

The force applied on it is again the gravity and this time its having a **positive acceleration** i.e. its following the same direction of the **acceleration due to gravity (g)**

#### velocity of the ball *just before touching the ground*

**And one important point**, during the downward fall the magnitude of the **velocity of the ball just before touching the ground** would be same as the magnitude of the velocity with which it was thrown upwards (v1 here). We will prove it mathematically here:

Here while falling vertically downward, the ball falls the same height H (here **H = v1 ^{2}/2g**, as given in equation iv)

During this downward displacement, the initial velocity is equal to the final velocity of the upward movement i.e. v2. And we know that v2=0.

And during the downward movement the final velocity is v3. This velocity is attained by the vertically falling ball just before touching the ground.

We have to find out the expression of this v3. Also we have to find out the time taken (say T) for this downward fall.

**v3 ^{2 }= v2^{2} + 2 g H** = 0 + 2 g (v1

^{2}/2g) = v1

^{2}

i.e.

**v3 = v1…………….(v)**

So we can say that during the downward fall the magnitude of the **velocity of the ball just before touching the ground** would be same as the magnitude of the velocity with which it was thrown upwards (v1 here).

### time for the downward movement when a ball is thrown vertically upward

It can be proved that the time for the downward movement or the time taken by the ball to fall from the highest point and reach the ground is same as the time required for the upward movement = v1/g

Let’s prove it here mathematically:

(see the diagram above for downward movememt)

**v3 = v2 + g T**

or, v3 = 0 + g T

so, **T = v3/g = v1/g** (*from equation v above*) ………….**(vi)**

**So Time for downward movement (T) = Time for upward movement ( t ) = v1/g**

That means, time for downward travel = time of upward travel

#### Total time required for the upward and downward movement

So (from equation ii and vi) for a vertically thrown object the **total time taken for its upward and downward movements** = t + T=**2v1/g ……….(vii)**

## A complete case study:

**Say a ball is thrown vertically upward with 98 m/s velocity, So v1 = 98 m/s**

1) The maximum height reached by it would be = **v1 ^{2}/2g= **(98 x98 )/(2 x 9.8) meter = 490 meter.

2) The time taken to reach the highest point = **v1/g** = 98 / 9.8 second = 10 second

3) The velocity at the highest point = 0

4) The time taken to reach the ground while falling from the highest point = **v1/g** = 98 / 9.8 second = 10 second

5) Total time taken for upward and downward movement = 10 sec + 10 sec = 20 sec.

6) Velocity just before touching the ground=same as initial velocity of throwing = 98 m/s

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