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# How to Find Maximum Height

## Introduction

Here, I will explain how to find maximum height in physics. I will also apply simple methods to solve a few problems about maximum height in projectile motion.

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.

The symbol for maximum height is Hmax. We write the maximum height formula of an object undergoing projectile motion as

Hmax = (U2Sin2θ) / 2g

Where

Hmax = 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 at2 which is changed into H = ut – ½ gt2 due to the effect of gravitational force

or apply

v2 = u2 + 2as which is changed into v2 = u2 – 2gh due to the effect of gravitational force

and H = Hmax = 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 v2 = u2 + 2as

when we substitute v = 0, a = -g and s = Hmax into the above equation, we now have

0 = u2 – 2gHmax

we now make take 2gHmax to the side of the zero to obtain

2gHmax = u2

By making the Hmax the subject of the formula, we now have

Hmax = u2 / 2g which is the formula for maximum height

For an angle θ

We have

Hmax = (u2sin2θ) / 2g where u = uy = usinθ

### Second Method of Deriving Maximum Height

Let us apply the second formula s = ut + ½ at2 to derive the maximum height

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

Hmax = ut – ½ gt2

Assuming u = ux = usinθ

since t = (usinθ) / g

we now have

Hmax = usinθ[(usinθ) / g] – ½ g[(usinθ) / g]2

Hmax = [u2sin2θ / g] – [(u2sin2θ) / 2g]

The above equation can further be written as

Hmax = [u2sin2θ / g] [1 – ½ ]

The above equation will now become

Hmax = [u2sin2θ / g] [½ ]

Therefore, we can finally write our maximum height as

Hmax = (u2sin2θ) / 2g

## Problems on How to Find Maximum Height

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

### Example 1

A ball is projected at an angle of elevation of 600 with an initial velocity of 100ms-1. Find the maximum height reached.

Solution

Data:

Initial velocity, u = 100ms-1

The angle of projection, θ = 600

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

The formula for maximum height, Hmax = (u2sin2θ) / 2g

and we can substitute our data into the above formula as

Hmax = (1002sin2600) / (2 x 10)

This implies

Hmax = (10,000 x sin2600) / (2 x 10) = (10,000 x 0.75) / 20 = 7,500/ 20 = 375m

### Example 2

A bullet is fired at an angle of 450 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 = 450

Initial velocity, u = 490 m/s

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

And the formula for maximum height, Hmax = (u2sin2θ) / 2g

we apply our data into the above formula

Hmax = (4902 x sin2450) / 2 x 10 = (240,100 x 0.5) / 20 = 120,050/20 = 6002.5 m = 6 km

### Example 3

A stone is shot out from a catapult with an initial velocity of 30 m/s at an elevation of 600. Find the maximum height attained.

Solution

Data:

initial velcoty, u = 30 m/s

angle of projection, θ = 600

gravitational acceleration, g = 10ms-2

Applying the formula for maximum height, Hmax = (u2sin2θ) / 2g

we now have

Hmax = (302 x sin2600) / 2 x 10

Therefore, it is now Hmax = (900 x 0.75) / 20 = 675/20 = 33.75 m

### Example 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, Hmax = u2 / 2g = 502 / 2 x 10 = 2500 / 20 = 125m

### Example 5

A bullet is fired from a point making an angel of 300 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 = 300

u = 40 m/s

g = 10 ms-2

Hmax = (u2sin2θ) / 2g

Hmax = (402 x sin2300) / 2 x 10 = (1,600 x 0.25) / 20 = 400/20 = 20 m

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Reference

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