# Horizontal Shot: Characteristics, Formulas And Equations, Exercises

The **horizontal shot** is the launching of a projectile with horizontal speed from a certain height and left to the action of gravity. Regardless of the air resistance, the path described by the mobile will have the shape of a parabola arc.

Projecting objects horizontally is quite common. The projectiles are launched for all kinds of purposes: from the stones with which the dams were struck at the beginning of the story, to those carried out in ball sports and are closely followed by crowds.

**characteristics**

The main characteristics of the horizontal shot are:

-The initial velocity given to the projectile is perpendicular to gravity.

-The movement takes place in a plane, so it takes two coordinates: *x* and *y.*

*–* It is done from a certain height H above ground level.

-The time the projectile lasts in the air is called the *flight time* .

-Factors such as air resistance or fluctuations in the value of *g* are not taken into account .

-The shape, size and mass of the projectile do not influence its movement.

-The movement is decomposed into two simultaneous movements: one vertical downwards under the action of *g* ; the other, horizontal, with constant speed.

__Formulas and equations__

__Formulas and equations__

The kinematic equations for horizontal launch are obtained from the equations for free fall and those for uniform rectilinear motion.

As the animation in Figure 1 clearly shows, the projectile is given a horizontal initial velocity, denoted as *v ** _{o} = v _{or }_{x } i* (bold in print indicates that it is a vector).

Note that the initial velocity has magnitude *v or _{x}* and is directed along the

*x-*axis , which is the direction of the unit vector

**. The animation also shows that the initial velocity does not have a vertical component, but as it falls, this component increases in a uniform way, thanks to the action of**

*i**g*, the acceleration of gravity.

As for the horizontal component of speed, it remains constant for the duration of the movement.

According to the aforementioned, the positions are established as a function of time, both on the horizontal axis and on the vertical axis. The direction to the right is taken as the + x axis, while the direction down is the -y direction. The value of gravity is *g = -9.8 m / s ^{2}* or

*-32 ft / s*:

^{2}*x (t) = x _{o} + v _{or }_{x} .t (Horizontal position); v or _{x} * is constant

*y (t) = y _{o} + v _{oy} .t – ½ gt ^{2} (vertical position); v _{y} = v _{oy} – gt (vertical speed)*

**Position, speed, time of flight and maximum horizontal range**

The equations are simplified by choosing the following starting positions: *x _{o} = 0* ,

*and*Furthermore

_{o}= 0 at the launch site.*v*, since the mobile is projected horizontally. With this choice the equations of motion are as follows:

_{oy}= 0*x (t) = v or _{x} .t; v _{x} = v _{ox}*

*y (t) = – ½ gt ^{2} ; v _{y} = – gt*

When time is not available, the equation relating velocities and displacements is useful. This is valid for the vertical speed, since the horizontal remains constant throughout the movement:

*v _{y }^{2} = v _{or y }^{2} + 2.g. y = 2.gy*

**Flight time**

To calculate the *flight time t _{flight}* , suppose that the mobile is projected from a height

*H*above the ground. As the origin of the reference system has been chosen at the launch point, when it reaches the ground it is in the

*–H*position . Substituting this in equation 2) we obtain:

*-H = – ½ gt ^{2 }_{flight}*

*t _{flight} = (2H / g) ^{½}*

**Maximum reach**

The *horizontal reach* is obtained by substituting this time in *x (t)* :

*x _{max}*

*= v*

_{ox}.*(2H / g)*

^{½}__Solved exercises__

__Solved exercises__

**-Solved exercise 1**

A helicopter flies horizontally, maintaining a constant elevation of 580 m when it drops a box containing food over a refugee camp. The box lands at a horizontal distance of 150 m from the point of its launch. Find: a) The flight time of the box.

b) The speed of the helicopter.

c) How quickly did the box land?

**Solution**

a) The height H from which the food is dropped is H = 500 m. With this data, when substituting, we obtain:

*t _{flight} = (2H / g) ^{½} = (2 x 580 / 9.8) ^{½} s = 10.9 s*

b) The helicopter carries the initial horizontal velocity *v or _{x}* of the packet and since one of the data is

*x*:

_{max}*x _{max}*

*= v*

_{ox}.*(2H / g)*®

^{½}*v or*

_{x}= x_{max}/ (2H / g)^{½}= x_{max}/ t_{flight}= 150 m / 10.9 s = 13.8 m / sc) The velocity of the projectile at any instant is:

*v _{y} = -gt = -9.8 m / s ^{2} x 10.9 s = -106.82 m / s = – 384.6 km / h*

The negative sign indicates that the mobile is moving downwards.

**-Solved exercise 2**

A package falls from an airplane flying horizontally at a height *H = 500 m* and *200 km / h and* must fall on an open vehicle traveling at *18 km / h* on the road. In what position must the plane drop the package so that it falls into the vehicle? Do not take into account air resistance or wind speed.

**Solution**

It is advisable to first pass all the units to the International System:

*18 km / h = 6 m / s*

*200 km / h = 55 m / s*

There are two mobiles: plane (1) and vehicle (2) and it is necessary to choose a coordinate system to locate them both. It is convenient to do it at the starting point of the package on the plane. The package is projected horizontally with the velocity that the plane carries: *v ** _{1}* , while the vehicle moves at

*v*_{ 2}assumed constant.

**-Airplane**

*Initial position: x = 0; y = 0*

*Initial velocity = v _{1} (horizontal)*

*Position equations: y (t) = -½g.t ^{2 } ; x (t) = v _{1} .t*

**-Vehicle**

*Initial position: x = 0, y = -H*

*Initial velocity = v _{2} (constant)*

*x (t) = x _{o} + v _{2} . t*

The duration of the package flight is:

*t _{flight} = (2H / g) ^{½}* =

*(2 × 500 / 9.8)*

^{½}s = 10.1 sIn this time, the package has experienced a horizontal displacement of:

*x _{max}*

*= v*

_{ox }.*(2H / g)*

^{½}= 55 m / sx 10.1 s = 556 m.In this time, the vehicle has moved horizontally as well:

*x (t) = v _{1} .t = 6 m / sx *

*10.1 s = 60.6 m*

If the plane drops the package immediately when it sees the vehicle passing under it, it will not make it fall right into it. For that to happen you must throw it further back:

*d = 556 m – 60.6 m = 495.4 m.*

__References__

__References__

- Bauer, W. 2011. Physics for Engineering and Sciences. Volume 1. Mc Graw Hill. 74-84.
- Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 1. Kinematics. Edited by Douglas Figueroa (USB). 117 – 164.
- Projectile Motion. Recovered from: phys.libretexts.org.
- Rex, A. 2011. Fundamentals of Physics. Pearson. 53-58.
- Tippens, P. 2011. Physics: Concepts and Applications. 7th Edition. McGraw Hill. 126-131.