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The photo has reduced the dimensions to 0.10 times the real dimensions. The strobe light flashed at 10.0 Hz. Ball A was released from rest.Ball B was given a horizontal push to the right.
What kind of motion is Ball A experiencing
Using Newton's Laws of Motion, please explain why Ball A's horizontal motion is what it is.
How long will it take Ball A to roll 5.60 x 102 cm if the ramp was that long?
Describe Ball B's horizontal motion.
Describe Ball B's vertical motion.
Using Newton's Laws of Motion, please explain why Ball B's horizontal motion is what it is.
How long will it take Ball B to roll 5.60 x 102 cm in the vertical direction if the ramp was that long?
How long will it take Ball B to roll 86 cm in the horizontal direction?A projectile is any object that is given some initial velocity by some means and continues in motion by its own inertia.
Once given this initial velocity, these objects are under the influence of no force other than gravity.
Without gravity, the object would follow a linear path; with gravity, the path curves.The curved path of a projectile is a combination of these horizontal and vertical components. The curved path of a ball (as shown) is best analyzed by considering the horizontal and vertical components of motion separately.
Projectiles move just like freely falling objects along the y-axis. The changing motion is due only to acceleration due to gravity.
If friction is ignored, then there are no forces acting on the x-axis. If the net force is zero (there are no forces which are unbalanced), then acceleration is zero. If acceleration is zero, then the object is either at rest or moving with constant velocity.
2-D Kinematics
Two-Dimensional Motion Investigation
The following stroboscopic photograph was made by releasing two balls, A and B. on a ramp covered with graph paper.
(For an interactive/zoomable version of the image at right, please navigate to: )
Ball A
Ball B
Projectile Motion
VERTICAL COMPONENT:
HORIZONTAL COMPONENT:
Horizontal Launch
If we worked with projectile questions in grade 11 they were horizontal launch problems.
We were new to motion, and new to working with motion in two dimensions.
Working with horizontal launches allowed us to ignore any components and combine a free-fall problem, with a constant velocity problem.
So, in the vertical (y) direction, we could use this:
dy=v1-y∆t+12ay∆t2
and since we have no initial vertical velocity…
dy=0+12ay∆t2
which makes finding ∆t really simple…
∆t=2∆dyay
and then in the horizontal (x) we would use this time, and the horizontal velocity.
dx=vx∆t
In our experiment we don’t know the time interval, so we’re going to have to do some substitution to find it/manage without it.
Horizontal Component
dx=vx∆t
Vertical Component
∆t=2∆dyay
Resulting in:
dx=vx2dyay
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