In Physics, various kinds of motions are studied. One such motion of any object is the Projectile motion. Projectile motion is one specific form of motion where the object moves in a bilaterally symmetrical, parabolic path. This path which the object follows is its trajectory. In this article, we will discuss the basic concept of projectile motion. Also, students will learn about many related computations in this motion. One such computation is the maximum height attained by that object. Here we will see the maximum height formula with examples. Let us learn it!

**Maximum Height in Projectile Motion**

A projectile is an object upon which throughout only one force acting i.e. due to gravity, except at the beginning. There are many examples of projectiles are there. An object dropped from the rest position is a projectile. Also, an object which is thrown vertically upward is a projectile, provided that the influence of air resistance is nowhere. And an object which is thrown upward at some angle with a horizontal plane is also a projectile.

**Some key points about this motion are:**

- Objects which are projected from and plane land on the same horizontal surface will always have a vertically symmetrical path.
- The time it takes from an object to be projected and the land is known as the time of flight. This time depends on the initial velocity of the projectile as well as on the angle of projection.
- When the object reaches a vertical velocity of zero magnitudes, then it will be at its maximum height of the projectile. Also, further gravity will take over and accelerate the object in a downward direction.
- The horizontal displacement of the object in the projectile is the range of the projectile, which will depend on the initial velocity of the object.

Source:en.wikipedia.org

**The Formula for Maximum Height**

The maximum height of the object in projectile motion depends on the initial velocity, the launch angle and the acceleration due to gravity. Its unit of measurement is “meters”. So Maximum Height Formula is:

\(Maximum \; height = \frac {(initial \; velocity)^2 (Sine \; of \; launch\; angle)^2}{2 \times acceleration\; due\; to \; gravity}\)

Mathematically: \(H = \frac {(v_0)^2 sin^2\theta }{2\times g}\)

- H is maximum height
- \(v_0\) is initial velocity per second
- g is acceleration due to gravity, i.e. \((9.80 m s^{-2})\)
- \(\theta\) is the angle of the initial velocity from the horizontal plane (radians or degrees)

**Solved Examples for Maximum Height Formula**

Q.1: A firefighter plane aims a fire hose upward, toward a fire in a skyscraper. The water leaving the hose with a velocity of 32.0 m per second. If the firefighter holds the hose at an angle of \(78.5 ^{\circ}\) Find out the maximum height of the water stream using maximum height formula.

Solution: The water droplets leaving the hose will be considered as the object in projectile motion. So its maximum height can be found using the said formula.

Now, given parameters are:

\(v_0 = 32 m per s\)

\(sin \theta = sin 78.5 ^{\circ} = 0.98\)

\(g = 9.8 m s^ {-2}\)

Thus, \(H = \frac {(v_0)^2 sin^2\theta }{2\times g} \\\)

\(= \frac {(32 )^2 \times (0.98) ^2}{2 \times 9.8} \\\)

\(= \frac {1024 \times 0.9604}{2 \times 9.8} \\\)

\(H \simeq 50.2 \;m\)

Thus the maximum height of the water from the hose will be 50.2 m.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…