## What is Maximum Height of a Projectile Motion?

**Maximum Height Definition:** The term “maximum height of a projectile motion” refers to the highest point reached by an object that is launched into the air and follows a curved path under the influence of gravity. It is the point where the object’s vertical velocity becomes zero before it begins descending back to the ground. In this article, I will walk you through a step-by-step method of how to find maximum height in physics. I will also apply simple methods to solve a few problems on this topic.

When we throw a stone, we apply an external force on it and it will travel along a parabolic path. It will also undergo the influence of gravitational force, and air resistance before it finally falls down. This is what we called a projectile, and the path it follows is called a trajectory.

## Understanding Maximum Height

When we throw an object vertically upward, it will accelerate vertically upward. The object will travel a certain distance (along the parabolic path) before it starts decelerating and finally falls to the ground. As the object is traveling along the parabolic paths, it follows a trajectory. We can describe the maximum height of an object as the highest vertical position attained by an object.

If you plot a graph to trace the motion and position of the object, you will observe that as soon as it reaches a certain height along the y-axis. The object will now start coming down to the ground.

Initially, when we throw an object vertically upward, it will accelerate along the trajectory. The acceleration will start reducing until the body reaches a certain height where it can no longer move upward again. The body will then start coming back to the ground after reaching it is peak height which we call maximum height.

## Maximum Height Formula

The symbol for maximum height is H_{max}. The maximum height formula of an object undergoing projectile motion is:

H_{max} = (U^{2}Sin^{2}θ) / 2g

Where

H_{max} = Maximum height

U = Initial velocity

θ = Angle of projection

g = Gravitational acceleration = constant = **9.8 ms ^{-2}** or

**10 ms**

^{-2}## How to Derive the Formula for Maximum Height

We can apply two of the equations of motion to derive the formula for maximum height. The two equations are as follows:

**s = ut + ½ at ^{2} which is changed into H = ut – ½ gt^{2} due to the effect of gravitational force**

or apply

**v ^{2} = u^{2} + 2as which is changed into v^{2} = u^{2} – 2gh due to the effect of gravitational force**

and H = H_{max} = maximum height

### First Method of Deriving Maximum Height

**If** **u is the initial velocity along the vertical component to reach the maximum height**

Let us apply the formula **v ^{2} = u^{2} + 2as**

when we substitute v = 0, a = -g and s = H_{max} into the above equation, we now have

0 = u^{2} – 2gH_{max}

we now make take 2gH_{max} to the side of the zero to obtain

2gH_{max} = u^{2}

By making the H_{max} the subject of the formula, we now have

**H _{max} = u^{2} / 2g** which is the formula for maximum height

For an angle θ

We have

**H _{max} = (u^{2}sin^{2}θ) / 2g** where

**u = u**

_{y}= usinθ### Second Method of Deriving Maximum Height

Let us apply the second formula ** s = ut + ½ at^{2}** to derive the maximum height

when we change s = H_{max}, a = -g (due to gravitational force). We will now obtain

**H _{max} = ut – ½ gt^{2}**

Assuming u = u_{x} = usinθ

since t = (usinθ) / g

we now have

H_{max} = usinθ[(usinθ) / g] – ½ g[(usinθ) / g]^{2}

H_{max} = [u^{2}sin^{2}θ / g] – [(u^{2}sin^{2}θ) / 2g]

The above equation can further be written as

H_{max} = [u^{2}sin^{2}θ / g] [1 – ½ ]

The above equation will now become

H_{max} = [u^{2}sin^{2}θ / g] [½ ]

Therefore, we can finally write our maximum height as

H_{max} = (u^{2}sin^{2}θ) / 2g

## Problems on How to Find Maximum Height

Here are solved problems to help you master how to find maximum height:

### Problem 1

A ball is projected at an angle of elevation of 60^{0} with an initial velocity of 100ms^{-1}. Find the maximum height reached.

**Solution**

**Data:**

Initial velocity, u = 100ms^{-1}

The angle of projection, θ = 60^{0}

gravitational acceleration is constant, and the symbol is g = 10ms^{-2}

The formula for maximum height, H_{max} = (u^{2}sin^{2}θ) / 2g

and we can substitute our data into the above formula as

H_{max} = (100^{2}sin^{2}60^{0}) / (2 x 10)

This implies

H_{max} = (10,000 x sin^{2}60^{0}) / (2 x 10) = (10,000 x 0.75) / 20 = 7,500/ 20 = 375m

### Problem 2

A bullet is fired at an angle of 45^{0} to the horizontal with a velocity of 490 m/s. Calculate the maximum height reached by the bullet.

**Solution**

**Data:**

The angle of projection, θ = 45^{0}

Initial velocity, u = 490 m/s

gravitational acceleration is constant, and the symbol is g = 10ms^{-2}

And the formula for maximum height, H_{max} = (u^{2}sin^{2}θ) / 2g

we apply our data into the above formula

H_{max} = (490^{2} x sin^{2}45^{0}) / 2 x 10 = (240,100 x 0.5) / 20 = 120,050/20 = 6002.5 m = 6 km

### Problem 3

A stone is shot out from a catapult with an initial velocity of 30 m/s at an elevation of 60^{0}. Find the maximum height attained.

**Solution**

**Data:**

initial velcoty, u = 30 m/s

angle of projection, θ = 60^{0}

gravitational acceleration, g = 10ms^{-2}

Applying the formula for maximum height, H_{max} = (u^{2}sin^{2}θ) / 2g

we now have

H_{max} = (30^{2} x sin^{2}60^{0}) / 2 x 10

Therefore, it is now H_{max} = (900 x 0.75) / 20 = 675/20 = 33.75 m

### Problem 4

A tennis ball is thrown vertically upwards from the front with a velocity of 50 m/s. Calculate the maximum height reached.

**Solution**

**Data:**

Initial velocity, u = 50m/s

gravitational acceleration, g = 10ms^{-2}

Also, H_{max} = u^{2} / 2g = 50^{2} / 2 x 10 = 2500 / 20 = 125m

### Problem 5

A bullet is fired from a point making an angel of 30^{0} to the horizontal, H. The initial velocity of the bullet is 40 m/s. Calculate the greatest height reached.

**Solution**

**Data:**

Angle of the bullet = 30^{0}

u = 40 m/s

g = 10 ms^{-2}

H_{max} = (u^{2}sin^{2}θ) / 2g

H_{max} = (40^{2} x sin^{2}30^{0}) / 2 x 10 = (1,600 x 0.25) / 20 = 400/20 = 20 m

### Problem 6

A stone is projected upwards at an angle of 30^{0} to the horizontal from the top of a tower of height 100m and it hits the ground at a point Q. If the initial velocity of projection is 100m/s, calculate the maximum height of the stone above the ground.

**Data: **The information from the question

Angle of projection, θ = 30^{0}

The maximum height of the stone above the ground, S, is equal to the maximum height, H from the point of projection and the height of the tower, 100m. Thus, S = 100 + H

Initial velocity, u = 100 m/s

We will now apply the formula: H_{max} = (u^{2}sin^{2}θ) / 2g

#### Solution

We will now substitute our formula with our data to obtain

H_{max} = (u^{2}sin^{2}θ) / 2g = (100^{2} x sin^{2}30^{0}) / (2 x 10) = 125m

However, we have S = 100 + H_{max} = 100 + 125 = 225 m

**Therefore, the maximum height of the stone above the ground is 225 meters. **

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How to Calculate Cubic Expansivity with Examples

**Reference**