Maximum height of a projectile formula

When an object is in flight in both horizontal and vertical directions and is influenced by gravity, then this motion is called projectile motion. The path traveled by the projectile is called the trajectory, which looks like a parabola. The horizontal distance covered by the projectile is called the range of the projectile. The two main components of the trajectory in a projectile motion are the distance covered (also called the projectile range) and the maximum height of the projectile. In this article, we derive the formula for the maximum height of the projectile and see some solved examples.

Why is the maximum height of a projectile important?

When any object is thrown, it follows a parabolic path. We are generally concerned about the distance covered by the object. But, for obtaining a certain amount of distance a certain maximum height is also necessary for the given initial speed. So, we have to balance the initial speed and the maximum height in order to obtain the desired distance or range. This becomes important in sports like javelin throw, high jump, pole vault, etc. This calculation is also important in the design of weapons like rockets, slingshots bows, and arrows, etc. For more examples of projectile motion, you can read our article on 20 examples of projectile motion.

Maximum Height of projectile motion

There are some points to be kept in mind while discussing the maximum height of a projectile motion.

  • The only force acting on the object in the projectile motion is the force of gravity with an acceleration of gravity g = ~9.8 m/s on Earth.
  • At the point of maximum height, the vertical component of velocity becomes zero.
  • All the 3 equations of motion are valid in a projectile motion.

Derivation for the formula for a maximum height of projectile motion

Derivation for the formula of maximum height of a projectile
Derivation for the formula of maximum height of a projectile

Using the third equation of motion:

V2 = u2 -2gs —(3)

  • The final velocity is zero here (v=0).
  • The initial velocity in the y-direction will be u*sinθ.
  • The displacement in the y-direction (S) will be the maximum height achieved by the projectile.

Using these values:

0 = u2 -2gs

S = (usinθ)2/2g 

Solved examples for the maximum height of a projectile

A person is participating in a high jump competition. Calculate the angle of jump the person has to take to jump across a height of 2 m. The initial speed of the person is 8 m/s.

The formula of maximum height of projectile is very important for sports like high jump.
In the sport of a high-jump, a person has to jump across a certain height(bar) without disturbing the bar.

Using the formula for a maximum height of projectile [S = (usinθ)2/2g]

2 = (8*sinθ)2/2*9.8

sin-1 (0.6125) = 37.7 degrees.

Can this jump be possible with a speed of 3m/s?

Again applying the same formula for maximum height,

2 = (3*sinθ)2/2*9.8

sin-1(4.35) = invalid

Hence the jump is not possible for a speed of 3m/s

Jimmy wants to throw an object into his house’s window, situated on the second floor(12m from the ground) from the ground. Calculate the speed required at an angle of 30.

Applying the formula for maximum height for a projectile [S = (usinθ)2/2g]

12 = (u*sin30)2/2*9.8

u = 30.672 m/s, which is a very high speed.

See also