# Horizontal shot: features, formulas and equations, exercises

The horizontal shot is the launching of a projectile with horizontal velocity from a certain height and left to the action of gravity. Regardless of air resistance, the path described by the cell phone will take the form of a parabola arc.

Projecting objects horizontally is quite common. Projectiles are thrown for all sorts of purposes: from the rocks with which dams were knocked down in early history to those that are used in ball sports and are closely followed by crowds

## Characteristics

The main features of horizontal shooting are:

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

-Motion takes place in a plane, so you need two coordinates: x and y.

It is made from a certain height H above ground level.

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

– Factors such as air resistance or fluctuations in the value of g are not considered .

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

-The movement is divided into two simultaneous movements: one vertical below the action of g ; the other, horizontal, with constant velocity.

## Formulas and Equations

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

As the animation in Figure 1 clearly shows, the projectile receives an initial horizontal velocity, denoted o = v ox i (bold in printed text indicates it is a vector).

Note that the initial velocity has magnitude ox and is directed along the x axis , which is the direction of the unit vector i . The animation also warns that the initial velocity has no vertical component, but as it falls, this component increases uniformly, thanks to the action of g , the acceleration of gravity.

As for the horizontal component of velocity, it remains constant during movement.

According to the above, positions are established as a function of time, both on the horizontal and vertical axis. The correct direction is taken as the +x axis, while the low position is the –y direction. The gravity value is g = -9.8 m / s 2 or -32 ft / s 2 :

x (t) = x or + v ox .t (horizontal position); v ox is constant

y (t) = y o + v oy .t – ½ gt 2 (vertical position); v y = v oy – gt (vertical velocity)

### Position, speed, flight time and maximum horizontal range

The equations are simplified if they choose the following starting positions: or = 0 , and or = 0 at the launch site. Also oy = 0 as the phone is projected horizontally. With this option, the equations of motion look like this:

x (t) = v ox. t; v x = v ox

and (t) = – ½ gt 2 ; v y = – gt

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

2 = v oy 2 + 2.g .y = 2.gy

#### Flight time

To calculate the flight time t flight , suppose the rover is projected from a height H above the ground. As the origin of the reference system at the starting point was chosen, when it reaches the ground, it is in the –H position . Substituting this into equation 2) you get:

-H = – ½ gt 2 of flight

flight = (2H / g) ½

#### range

The horizontal range is obtained by replacing this time in x (t) :

max = v ox . (2H / g) ½

## solved exercises

### Exercise -Fixed 1

A helicopter flies horizontally, maintaining a constant elevation of 580 m when releasing a crate containing food into 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 food is released is H = 500 m. With this data when replacing, you get:

flight = (2H / g) ½ = (2 x 580 / 9.8) ½ s = 10.9 s

b) The helicopter carries the initial horizontal velocity bo of the package and, since one of the data is max :

max = v ox . (2H / g) ½ ® ox = x max / (2H / g) ½ = x max / t flight = 150 m / 10.9 s = 13.8 m / s

c) The velocity of the projectile at any time is:

y = -gt = -9.8 m / s 2 x 10.9 s = -106.82 m / s = – 384.6 km / h

### – Exercise solved 2

From an airplane flying horizontally at a height H = 500 m and 200 km/h a package that can fall onto an open vehicle descends 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

All units must be transferred to the International System first:

18 km / h = 6 m / s

200 km / h = 55 m / s

There are two cell phones: plane (1) and vehicle (2) and you need to choose a coordinate system to locate both. It is convenient to do this at the starting point of packaging on the plane. The package is projected horizontally at the speed the plane carries: 1 , while the vehicle moves to supposedly constant v 2 .

-Plane

Starting position: x = 0; y = 0

Initial velocity = 1 (horizontal)

Position equations: y (t) = -½g.t 2 ; x (t) = v 1 .t

-Vehicle

Starting position: x = 0, y = -H

Initial velocity = 2 (constant)

x (t) = x or + v 2 . t

The duration of the package flight is:

flight = (2H / g) ½ = (2 × 500 / 9.8) ½ s = 10.1 s

During this period, the package suffered a horizontal displacement of:

max = v ox . (2H / g) 1/2 = 55 m / sx 10.1 s = 556 m.

During this period, the vehicle also moved horizontally:

x (t) = v 1 .t = 6 m / sx 10.1 s = 60.6 m

If the plane drops the package immediately where the vehicle is passing under it, it will not cause it to fall straight into it. For that to happen, you must throw it backwards:

d = 556 m – 60.6 m = 495.4 m.

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