4.3 Projectile motion (Page 2/13)

University physics volume 1 Page 2 / 13

Horizontal Motion

v_{0 x} = v_{x}, x = x_{0} + v_{x} t

Vertical Motion

y = y_{0} + \frac{1}{2} (v_{0 y} + v_{y}) t

v_{y} = v_{0 y} - g t

y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2}

v_{y}^{2} = v_{0 y}^{2} - 2 g (y - y_{0})

Using this set of equations, we can analyze projectile motion, keeping in mind some important points.

Problem-solving strategy: projectile motion

Resolve the motion into horizontal and vertical components along the x - and y -axes. The magnitudes of the components of displacement $\vec{s}$ along these axes are x and y. The magnitudes of the components of velocity $\vec{v}$ are $v_{x} = v cos θ and v_{y} = v sin θ,$ where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in [link] .
Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
Recombine quantities in the horizontal and vertical directions to find the total displacement $\vec{s}$ and velocity $\vec{v} .$ Solve for the magnitude and direction of the displacement and velocity using

$s = \sqrt{x^{2} + y^{2}}, θ = {tan}^{−1} (y / x), v = \sqrt{v_{x}^{2} + v_{y}^{2}},$

where θ is the direction of the displacement $\vec{s} .$

Figure a shows the locations and velocities of a projectile on an x y coordinate system at 10 instants in time. When the projectile is at the origin, it has a velocity v sub 0 y which makes an angle theta sub 0 with the horizontal. The velocity is shown as a dark blue arrow, and its x and y components are shown as light blue arrow. The projectile’s position follows a downward-opening parabola, moving up to a maximum height, then back to y = 0, and continuing below the x axis The velocity, V, at each time makes an angle theta which changes in time, and has x component V sub x and y component v sub y. The x component of the velocity V sub x is the same at all times. The y component v sub y points up but gets smaller, until the projectile reaches the maximum height, where the velocity is horizontal and has no y component. After the maximum height, the velocity has a y component pointing down and growing larger. As the projectile reaches the same elevation on the way down as it had on the way up, its velocity is below the horizontal by the same angle theta as it was above the horizontal on the way up. In particular, when it comes back to y = 0 on the way down, the angle between the vector v and the horizontal is minus s=theta sub zero and the y component of the velocity is minus v sub 0 y. The last position shown is below the x axis, and the y component of the velocity is larger than it was initially. The graph clearly shows that the horizontal distances travelled in each of the time intervals are equal, while the vertical distances decrease on the way up and increase on the way down. Figure b shows the horizontal component, constant velocity. The horizontal positions and x components of the velocity of the projectile are shown along a horizontal line. The positions are evenly spaced, and the x components of the velocities are all the same, and point to the right. Figure c shows the vertical component, constant acceleration. The vertical positions and y components of the velocity of the projectile are shown along a vertical line. The positions are get closer together on the way up, then further apart on the way down. The y components of the velocities initially point up, decreasing in magnitude until there is no y component to the velocity at the maximum height. After the maximum height, the y components of the velocities point down and increase in magnitude. Figure d shows that putting the horizontal and vertical components of figures b and c together gives the total velocity at a point. The velocity V has an x component of V sub x, has y component of V sub y, and makes an angle of theta with the horizontal. In the example shown, the velocity has a downward y component. — (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because $a_{x} = 0$ and $v_{x}$ is a constant. (c) The velocity in the vertical direction begins to decrease as the object rises. At its highest point, the vertical velocity is zero. As the object falls toward Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x and y motions are recombined to give the total velocity at any given point on the trajectory.

A fireworks projectile explodes high and away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of

75.0 °

above the horizontal, as illustrated in [link] . The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d) What is the total displacement from the point of launch to the highest point?

The trajectory of a fireworks shell from its launch to its highest point is shown as the left half of a downward-opening parabola in a graph of y as a function of x. The maximum height is h = 233 meters and its x displacement at that time is x = 125 meters. The initial velocity vector v sub 0 is up and to the right, tangent to the trajectory curve, and makes an angle of theta sub 0 equal to 75 degrees. — The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Strategy

The motion can be broken into horizontal and vertical motions in which

a_{x} = 0

and

a_{y} = − g .

We can then define

x_{0}

and

y_{0}

to be zero and solve for the desired quantities.

Solution

(a) By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex , is reached when

v_{y} = 0 .

Since we know the initial and final velocities, as well as the initial position, we use the following equation to find y :

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Practice Key Terms 4

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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