Question

How do you determine the maximum height of a projectile?

Answer 1

Projectile motion:

1. The motion of an object that is thrown in the air, subjected to only the acceleration due to gravity, is known as projectile motion.
2. First, the body goes up due to the force applied while throwing and then gradually comes down due to acceleration due to gravity.
3. The object that is thrown is called a projectile, and the path that it takes is called a trajectory.

Maximum height of a projectile:

1. The maximum height of a projectile is given by the formula $H=\frac{{\left(u\sin \theta \right)}^2}{2g}$, where $u$ is the initial velocity, $\theta$ is the angle at which the object is thrown and $g$ is the acceleration due to gravity.
2. This is derived by using the third equation of motion $v^2=u^2-2gS$, where $v$ is the final velocity, $u$ is the initial velocity, $g$ is the acceleration due to gravity and $S$ is the distance covered.
3. Here, the final velocity is $0$, since the body comes to a rest, the initial velocity is $u\sin \theta$ where $\theta$ is the angle at which the object is thrown and the distance travelled is equal to the maximum height $H$. So, the equation $v^2=u^2-2gS$ can be written as $0^2={\left(u\sin \theta \right)}^2-2gH$.
4. Thus, the maximum height of a projectile is $H=\frac{{\left(u\sin \theta \right)}^2}{2g}$.
5. The diagram below describes the projectile motion and its maximum height. If a body is thrown into the air at an angle, its velocity has a horizontal component $v_x$ and a vertical component $v_y$.

Hence, the maximum height of a projectile is $H=\frac{{\left(u\sin \theta \right)}^2}{2g}$.