![]() The first solution is the starting time, and the second is the time at which the toy rocket reaches the ground again. There are two solutions for t, depending whether the positive or negative in the ± symbol is chosen. This has the form of the quadratic equation, with t as the variable: ![]() So, set the left side of the equation equal to that height: ![]() The field is flat, so the rocket will hit the ground again at y = 0.00 m. To find how long the toy rocket was in the air, use the equation for the vertical distance for the rocket: So, the initial velocity in the x direction v xo is:Īnd the initial velocity in the y direction is: So, the components of the velocity can be set equal to the parts of the equation above: The equation for vector addition of the initial velocity components is: We multiply both sides by the initial velocity squared v o 2: The initial velocity can be broken down using an equation relating the sine and cosine: The velocity that is given has both x and y components, because it is in a direction 60.0° up from the horizontal (x) direction. How much time does the toy rocket spend the air, and how far from its launch point does it land in the field?Īnswer: The first thing that must be found to solve this problem is the initial velocity in the x and y directions. Its initial velocity has a magnitude of 20.0 m/s. Therefore, the velocity (magnitude and direction) of the ball after 5.00 s was 51.24 m/s, -72.98° down from the horizontal.Ģ) A toy rocket is launched in a flat field, aimed at an angle 60.0° up from the horizontal (x) axis. If the horizontal direction is 0.0 radians, the angle can be found with the equation: Though it was not asked for in the question, it is also possible to find the direction of the velocity as an angle. The magnitude of the velocity is 51.24 m/s. To find the magnitude of the velocity, the x and y components must be added with vector addition: In projectile motion problems, up is defined as the positive direction, so the y component has a magnitude of 49.0 m/s, in the down direction. The x component of the velocity after 5.00 s is: The ball was kicked horizontally, so v xo = 15.0 m/s, and v yo = 0.0 m/s. Once these two components are found, they must be combined using vector addition to find the final velocity. After 5.00 s, what is the magnitude of the velocity of the ball?Īnswer: The velocity of the ball after 5.00 s has two components. The initial velocity of the ball is 15.0 m/s horizontally. G = acceleration due to gravity (9.80 m/s 2)ġ) A child kicks a soccer ball off of the top of a hill. ![]() Vertical velocity = initial vertical velocity - (acceleration due to gravity)(time) Horizontal velocity = initial horizontal velocity Horizontal distance = (initial horizontal velocity)(time) The horizontal and vertical velocities are expressed in meters per second (m/s). The units to express the horizontal and vertical distances are meters (m). Velocity is a vector (it has magnitude and direction), so the overall velocity of an object can be found with vector addition of the x and y components: v 2 = v x 2 + v y 2. The trajectory has horizontal (x) and vertical (y) components. The path the object follows is determined by these effects (ignoring air resistance). To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical.A projectile is an object that is given an initial velocity, and is acted on by gravity. Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly. You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Figure 5.28 compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). The most important concept in projectile motion is that horizontal and vertical motions are independent, meaning that they don’t influence one another. Air resistance does significantly alter trajectory motion, but due to the difficulty in calculation, it is ignored in introductory physics. As an object travels through the air, it encounters a frictional force that slows its motion called air resistance. The object is called a projectile, and its path is called its trajectory. After the initial force that launches the object, it only experiences the force of gravity. Projectile motion is the motion of an object thrown (projected) into the air.
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