Objects that are launched into the air are called projectiles. The path followed by a projectile in motion is called a trajectory. The image of the snowboarder in the chapter introduction showed his trajectory; every object has a trajectory even when we cannot see it. Projectile motion, or the object's trajectory, is described in terms of position, velocity, and acceleration. Even with only an object's current location, velocity, and acceleration, we can calculate when and where the object will land. Our knowledge that perpendicular components of vectors do not affect each other allow us to easily analyze the motion of projectiles.
Diagram of two balls (one red and one blue) dropped at the same time, with a photograph taken every second so there are 5 images of the two balls
In the diagram, two balls (one red and one blue) are dropped at the same time. The red ball is released with no horizontal motion and the blue ball is dropped but also given a horizontal velocity of 10 m/s. As the balls fall to the floor, a photograph is taken every second so that in 5 seconds, we have 5 images of the two balls. Each vertical line on the diagram represents 5 m. Since the blue ball has a horizontal velocity of 10 m/s, you will see that for every second, the blue ball has moved horizontally 10 m. That is, in each second, the blue ball has increased its horizontal distance by 10 m. This horizontal motion is due to the ball's constant velocity.
The red ball was dropped straight down with no horizontal velocity and therefore, in each succeeding second, the red ball falls straight down with no horizontal motion. The increasing distances between seconds in the red ball's motion indicates that this motion is accelerating.
A very important point here is that the vertical motion of these two balls is identical. That is, they each fall exactly the same distance vertically in each succeeding second. The constant horizontal velocity of the blue ball has no effect on its accelerated vertical motion. Therefore, the vertical motion of the blue ball can be analyzed exactly the same as the vertical motion of the red ball.
Examples
Example 1
If an arrow is fired from a bow with a perfectly horizontal velocity of 60.0 m/s and the arrow was 2.00 m above the ground when the it was released, how far will the arrow fly horizontally before it strikes the ground?
This problem is solved by determining how long it takes the arrow to fall to the ground in exactly the same manner as if the arrow was dropped with no horizontal velocity. The time required for the arrow to fall to the ground will be the same time that the arrow flies horizontally at 60.0 m/s, so
d=12at2 solved for t=2da−−−√=(2)(2.00 m)9.80 m/s2−−−−−−−−−−√=0.639 s
The time required for the fall is multiplied by the horizontal velocity to get the horizontal distance.
dhorizontal=(vhorizontal)(time)=(60.0 m/s)(0.639 s)=38.3 m
Example 2
A rock was thrown horizontally from a 100.0 m high cliff. It strikes the ground 90.0 m from the base of the cliff. At what speed was it thrown?
We can calculate how long it takes for a rock to free fall 100.0 m and then divide this time into the horizontal distance to get the horizontal velocity.
tv=2da−−−√=(2)(100.0 m)9.80 m/s2−−−−−−−−−−−√=4.52 s=dt=90.0 m4.52 s=19.9 m/s
Use the following simulation to adjust the initial velocity and height of a launched arrow and try to predict whether or not the archer will hit her target:
Explanation:
