Which projectile traveled the greatest horizontal distance

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. Example A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of The fuse is timed to ignite the shell just as it reaches its highest point above the ground. The highest point in any trajectory, called the apex , is reached when.

Since we know the initial and final velocities, as well as the initial position, we use the following equation to find y :. Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative.

Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding.

In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height. In this case, the easiest method is to use. This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes.

Another way of finding the time is by using. The horizontal displacement is the horizontal velocity multiplied by time as given by. Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators.

When the shell explodes, air resistance has a major effect, and many fragments land directly below. Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure , which shows the curvature of the trajectory toward the ground level. When solving Figure a , the expression we found for y is valid for any projectile motion when air resistance is negligible.

This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity. On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit.

Again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solve for t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain. We can find the time for this by using Figure :.

The initial vertical velocity is the vertical component of the initial velocity:. Substituting into Figure for y gives us. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator:. The time for projectile motion is determined completely by the vertical motion.

Thus, any projectile that has an initial vertical velocity of Then, we can combine them to find the magnitude of the total velocity vector. We choose the starting point because we know both the initial velocity and the initial angle. The final vertical velocity is given by Figure :. This result is consistent with the fact that the ball is impacting at a point on the other side of the apex of the trajectory and therefore has a negative y component of the velocity.

The magnitude of the velocity is less than the magnitude of the initial velocity we expect since it is impacting Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface.

In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections. We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find.

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Figure does not apply when the projectile lands at a different elevation than it was launched, as we saw in Figure of the tennis player hitting the ball into the stands.

The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long. The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y x.

We take. Substituting the expression for t into the equation for the position. From the trajectory equation we can also find the range , or the horizontal distance traveled by the projectile. Factoring Figure , we have. The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontal surface.

Note particularly that Figure is valid only for launch and impact on a horizontal surface. We see the range is directly proportional to the square of the initial speed. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we see from the factor. These results are shown in Figure. In a we see that the greater the initial velocity, the greater the range.

In b , we see that the range is maximum at. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to. The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Note that the range is the same for initial angles of. Example Comparing Golf Shots A golfer finds himself in two different situations on different holes. On the second hole he is m from the green and wants to hit the ball 90 m and let it run onto the green. He angles the shot low to the ground at.

On the fourth hole he is 90 m from the green and wants to let the ball drop with a minimum amount of rolling after impact. The previous diagrams, tables, and discussion pertain to how the horizontal and vertical components of the velocity vector change with time during the course of projectile's trajectory. Now we will investigate the manner in which the horizontal and vertical components of a projectile's displacement vary with time. As has already been discussed , the vertical displacement denoted by the symbol y in the discussion below of a projectile is dependent only upon the acceleration of gravity and not dependent upon the horizontal velocity.

Thus, the vertical displacement y of a projectile can be predicted using the same equation used to find the displacement of a free-falling object undergoing one-dimensional motion. This equation was discussed in Unit 1 of The Physics Classroom. The equation can be written as follows. The above equation pertains to a projectile with no initial vertical velocity and as such predicts the vertical distance that a projectile falls if dropped from rest.

It was also discussed earlier , that the force of gravity does not influence the horizontal motion of a projectile. The horizontal displacement of a projectile is only influenced by the speed at which it moves horizontally v ix and the amount of time t that it has been moving horizontally.

Thus, if the horizontal displacement x of a projectile were represented by an equation, then that equation would be written as. The diagram below shows the trajectory of a projectile in red , the path of a projectile released from rest with no horizontal velocity in blue and the path of the same object when gravity is turned off in green.

The position of the object at 1-second intervals is shown. As can be seen in the diagram above, the vertical distance fallen from rest during each consecutive second is increasing i. Furthermore, since there is no horizontal acceleration, the horizontal distance traveled by the projectile each second is a constant value - the projectile travels a horizontal distance of 20 meters each second.

Thus, the horizontal displacement is 20 m at 1 second, 40 meters at 2 seconds, 60 meters at 3 seconds, etc. This information is summarized in the table below. Now consider displacement values for a projectile launched at an angle to the horizontal i. How will the presence of an initial vertical component of velocity affect the values for the displacement? The diagram below depicts the position of a projectile launched at an angle to the horizontal.

The projectile still falls 4. These distances are indicated on the diagram below. The projectile still falls below its gravity-free path by a vertical distance of 0. Q: Light from two sources passes through a circular aperture to form images on a screen.

State the Rayl A: If Light from two sources passes through a circular aperture and forms image on a screen then, The R A: When the nucleus fission into two nuclei of same masses, their nucleus fission reaction can be writt Calculate the moment of inertia of a hollow cylinder of mass M and length - with inner radius a a A: Given: The mass of the cylinder is M. The length of the cylinder is L.

The inner radius is a. The ou Q: In oscillating water column, Ammonia used as the working fluid. Select one: True False. Q: What is the phase constant for the harmonic oscillator with the velocity function v t described in Q: 6- In below figure, find the equivalent spring constant of the system in the direction of 0.

Q: The switch in Fig. Q2 has been in position a for a long time. Q: 12— Q: A hydrogen atom is 0. Use the uncertainty principle to estimate the minimum e Q: b Cf undergoes alpha decay. A: Here the process to be undergone is Alpha decay The reaction involved in this process will be The el A second coil wit A: Given Data, Two long solenoid one has A: Given, An underdamped harmonic oscillator.

Q: Speed of Earth Verify that the length of one orbit of Earth isapproximately 6. A: Given: To Verify that the length of one orbit of Earth is approximately 6. To determine the Q: A kicking-simulation attachment goes on the front of a wheelchair, allowing athletes with mobility i Q: In photobromination of cinnamic acid, radiation at