Level I: If a rider accidentally releases an object when the ride is at maximum height and top speed, how far away
(horizontally) would that object land, assuming negligible air resistance? If viewed from above, what would the path of that
object look like?
To find the horizontal distance an object would land when released from a ride at maximum height and top speed, we can use the principle of projectile motion.
Projectile motion is the motion of an object thrown or projected into the air, under the force of gravity, with a given initial velocity and launch angle. In this case, we assume that the object is launched horizontally.
When the object is released, it will have an initial horizontal velocity component equal to the velocity of the ride. This horizontal velocity will remain constant throughout the motion since there are no horizontal forces acting on the object, neglecting air resistance.
The vertical velocity component will be zero when the object is released since it was initially moving horizontally. However, gravity will act on the object, and the vertical displacement of the object will follow a standard free fall motion.
The formula to calculate the horizontal distance traveled by the object is:
horizontal distance = horizontal velocity x time
Since the vertical displacement of the object can be calculated using the formula for free fall:
vertical displacement = (1/2) x acceleration due to gravity x time^2
And we know that time of flight (the time taken for the object to reach the ground) can be calculated using:
time of flight = 2 x (maximum height / acceleration due to gravity)^0.5
Therefore, using these equations, we can find the horizontal distance the object will travel.
As for the path of the object when viewed from above, it will follow a projectile motion trajectory, which is a parabolic path. The initial horizontal velocity will keep the object moving horizontally, while gravity will cause the object to move vertically in a symmetrical manner.