This chart shows the parabolic path of a ball rolled off of something like a table and allowed to fall freely. Click the image to set it in motion.
Notice that the forward velocity is constant, while the downward motion is accelerated. You can run the animation a few times. Look at both the forward progress (top) and the downward progress (left) of the ball.
The essence of how we treat projectile motion, the motion of a launched object after no more launching forces are working on it, is in this figure: The horizontal velocity (ignoring friction) is constant, and the vertical acceleration is just that of a freely-falling object.
A projectile is any object set free of any forces except for gravity and friction. A projectile can be a thrown ball, a bullet or a springboard diver ...
Except for air resistance, the forward velocity of any projectile is constant and is equal to the initial velocity when it was released. The vertical velocity changes by the acceleration of gravity.
Nature doesn't care that a projectile is moving forward. It's downward acceleration is just the acceleration of gravity, exactly as though it was dropped or thrown straight upward.
Consider the forces at work on a ball while it's rolling along the table. We assume that a force was or is being applied that created a constant velocity vx. The force of gravity, Fg, on the ball is exactly balanced by the normal force, FN, of the table pushing back on the ball.
Now as the ball rolls off the table, the normal force vanishes and the force of gravity pulls it downward. Unbalanced forces produce acceleration, so the ball will accelerate downward at 9.8 m/s2. There is no force acting horizontally (except for friction – air resistance), so the forward velocity will remain vx
As the ball falls, two things are going on independently. First, the ball is traveling forward at a constant velocity. You can see that by the even spacing of the tick marks along the horizontal axis in the figure below.
As time unfolds, the downward velocity of the ball increases at the rate of 9.8 m/s2, creating the uneven spacing in the vertical location of the ball (ticks on vertical axis).
That's the essence of projectile motion, no matter how complicated the scenario: Nature doesn't care about whether a projectile is moving horizontally. It's still going to be acted on by the force of gravity just as though it were dropped or thrown straight upward.
Solution: Here's a picture of the problem →
When the ball rolls off the edge of the table, it will continue moving forward at 2.0 m/s until it hits the floor. So we just need to find out how long it will be in the air before it hits.
For that, we'll use the freefall formula, rearranged to give us the time. What's crucial here is to realize that, when it comes to the vertical dimension, the ball is acting just like a dropped ball. Nature doesn't care that it's also moving forward at the same time.
Plugging in the height of the table and the acceleration of gravity, we get the time in seconds:
So the ball will be in the air (and thus able to move forward) for a little more than half a second. Now we just rearrange the speed equation,
and plug in the speed and time to get the distance traveled:
The ball will travel about 1.1 meters before hitting the ground.
Projectile motion problems always have this kind of two-step structure (at least).
Here's a picture of the situation. The ball is hit upward at a 45˚ angle with the ground, but gravity takes over immediately, bending the path into the downward parabola shown. For this problem we'll assume the distance from the bat to the ground is too small to matter.
We'll need to break that initial velocity vector into its vertical and horizontal components to get the vertical velocity, vy and the horizontal velocity vx. To do that we can use the 45-45-90 triangle, which you should memorize.
The length of a side of the 45-45-90 triangle is √2/2 times the length of the hypotenuse, so our hitter diagram looks like this:
Now to find the total time in the air, we use the definition of acceleration to find the time it takes the ball to get to the top of its arc, the acceleration in this case being g = 9.8 m/s2. We can rearrange that definition to find the elapsed time.
We are able to calculate the time because the velocity at the top of the arc, vf, is zero. The one-way time is:
The round trip time – up and down – is twice that, 5.772 s. Now it's just a matter of calculating the distance traveled during that time, given that the horizontal velocity is 20√2 m/s.
This is a pretty typical projectile motion problem that you should know how to solve.
Remember that this distance is really an upper limit to the distance that the ball could travel. In reality, air resistance would slow its travel a bit and the distance would decrease.
Solution: In this problem we'll examine the effect of launch angle on the range (distance traveled) by a projectile. Here's the picture:
The question we're really asking here is, which angle is the best for achieving the largest distance of the shot? We'll guess that it's 45˚ and test an angle on either side, 40˚ and 50˚.
This problem will require a little trigonometry because not all of those angles belong to convenient memorized triangles. The vertical (vy) and horizontal (vx) components of the muzzle velocity (that's just armament talk for the velocity of the cannonball or bullet as it exits the barrel) are shown here:
Let's calculate those vector components of the velocity for each angle using the sines and cosines:
Now we need to calculate the total time in the air using the definition of acceleration and the fact that the vertical velocity at the top of the arc is zero. The factor of 2 in this equation accounts for the round-trip up and back down.
Here are the results for each angle:
Now to calculate the distances using the horizontal velocities and the time interval,
Finally, here are the distances:
Indeed, the 45˚ angle yields the greatest distance. This is always true for a given initial velocity, all other conditions being equal.
Solution: First we need a diagram (I can't overstate the importance of a diagram for organizing your work in all kinds of physics problems).
In this problem, we'll be a little more precise and take the height of the muzzle of the cannon into account, too.
The first step is to calculate vertical and horizontal components of the initial velocity, vy and vx:
Now to calculate the distance and time to the top of the arc. The total time will have to be calculated in two steps because the up and down distances are different.
We'll need the upward distance in order to calculate the time to the bottom of the cliff.
The velocity in that last step is the average velocity of the climb, the average of 90.11 m/s and 0.0 m/s. Now the time it takes for the projectile to fall is calculated from a rearranged freefall equation and by adding 50 m + 1.5 m to the upward distance:
Now the total time that the projectile spends moving forward (before it hits the ground) is
Finally, as always, we use the horizontal velocity and the total time to calculate the distance traveled:
That's pretty far. The cliff gives us an advantage. Can you think of a simple change that would give us even more distance?
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