Zero Launch Angle

An object launched horizontally at a height $H$ travels a range $v_0 \sqrt{\frac{2H}{g}}$ during a time of flight $T = \sqrt{\frac{2H}{g}}$ .

Learning Objective

Explain the relationship between the range and the time of flight

Key Points

For the zero launch angle, there is no vertical component in the initial velocity.
The duration of the flight before the object hits the ground is given as T = \sqrt{\frac{2H}{g}} .
In the horizontal direction, the object travels at a constant speed v₀ during the flight. The range R (in the horizontal direction) is given as: $R= v_0 \cdot T = v_0 \sqrt{\frac{2H}{g}}$ .

Term

trajectory
The path of a body as it travels through space.

Examples

Lets go through the same example we did in the last section. An object is launched at an initial speed of 10 m/s. What is the range, maximum height and time of flight of the object at a 90 degree launch angle? We already know that with a 90 degree launch angle, the range is going to be zero, so no need to work anything out there.
Height: $\displaystyle {h=\frac{v_i^2}{2g}\\ h=\frac{10(\frac{\text{m}}{\text{s}})^2} {2\cdot 9.81(\frac{\text{m}}{\text{s}})}\\ h=\frac{100(\frac{\text{m}}{\text{s}})^2}{19.62(\frac{\text{m}}{\text{s}^2})} =5.10 m\\ }$ Time of Flight: $\displaystyle{T=\frac{2\cdot v_i}{g}\\ T=\frac{2\cdot 10(\frac{\text{m}}{\text{s}})}{9.81(\frac{\text{m}}{\text{s}^2})}\\ T=\frac{20(\frac{\text{m}}{\text{s}})}{9.81(\frac{\text{m}}{\text{s}^2})} =2.04 \text{s}\\ }$

Full Text

Projectile motion is a form of motion where an object moves in a parabolic path. The path followed by the object is called its trajectory. Projectile motion occurs when a force is applied at the beginning of the trajectory for the launch (after this the projectile is subject only to the gravity).

One of the key components of the projectile motion, and the trajectory it follows, is the initial launch angle. The angle at which the object is launched dictates the range, height, and time of flight the object will experience while in projectile motion. shows different paths for the same object being launched at the same initial velocity and different launch angles. As illustrated by the figure, the larger the initial launch angle and maximum height, the longer the flight time of the object.

Projectile Trajectories

The launch angle determines the range and maximum height that an object will experience after being launched.This image shows that path of the same object being launched at the same speed but different angles.

We have previously discussed the effects of different launch angles on range, height, and time of flight. However, what happens if there is no angle, and the object is just launched horizontally? It makes sense that the object should be launched at a certain height ($H$), otherwise it wouldn't travel very far before hitting the ground. Let's examine how an object launched horizontally at a height $H$ travels. In our case is when $\alpha$ is 0.

Projectile motion

Projectile moving following a parabola.Initial launch angle is $\alpha$, and the velocity is $v_0$.

Duration of Flight

There is no vertical component in the initial velocity ($v_0$) because the object is launched horizontally. Since the object travels distance $H$ in the vertical direction before it hits the ground, we can use the kinematic equation for the vertical motion:

$(y-y_0) = -H = 0\cdot T - \frac{1}{2} g T^2$

Here, $T$ is the duration of the flight before the object its the ground. Therefore:

$\displaystyle T = \sqrt{\frac{2H}{g}}$

Range

In the horizontal direction, the object travels at a constant speed $v_0$ during the flight. Therefore, the range $R$ (in the horizontal direction) is given as:

$\displaystyle R= v_0 \cdot T = v_0 \sqrt{\frac{2H}{g}}$

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