Question

A 100 g particle experiences the one-dimensional.Suppose the particle is shot toward the right from x = 1.0 m with a speed of 22 m/s . Where is the particle's turning point? Express your answer with the appropriate units.

Answer 1

The particle's turning point is located at x = 25.7 m to the right of its initial position of x = 1.0 m.

Step-by-step explanation:

To find the particle's turning point, we need to determine when the particle comes to a stop. Since the particle is shot toward the right, it will decelerate due to gravitational force until it eventually reaches a turning point and starts moving back towards the left. To find the turning point, we can use the kinematic equation vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the displacement. In this case, since the particle comes to a stop, the final velocity is 0 m/s.

Using the given information, we have vi = 22 m/s, a = -9.8 m/s^2 (acceleration due to gravity), and vf = 0 m/s. Plugging these values into the equation, we can solve for d:

0^2 = 22^2 + 2(-9.8)d

484 = 19.6d

d = 24.7 m

Therefore, the particle's turning point is located at x = 1.0 m + 24.7 m = 25.7 m. The particle's turning point is 25.7 meters to the right of its initial position.

Answer 2

To find the turning point of a particle shot horizontally, one would typically look for changes in the vertical direction due to gravity. However, since the particle in the question is shot horizontally with no initial vertical component and no other forces acting, its turning point would be where it impacts the ground, which isn't specified in the question.

Step-by-step explanation:

The question is about determining the turning point of a particle that is moving under the influence of gravity after being shot horizontally. When an object is thrown or shot horizontally, the only acceleration acting on it (neglecting air resistance) is due to gravity, which is downward. Therefore, the horizontal component of its motion will not change, meaning it will continue to move to the right at the same speed, in this case, 22 m/s. The vertical component of its motion, however, will be affected by gravity, which will act to slow down its upward motion (if any), and eventually reverse it, causing the object to fall back down. In this problem, however, we do not have an initial vertical velocity component, so the vertical motion is purely the acceleration due to gravity from the start.

In this problem, the 100 g particle is shot horizontally, thus it has no initial vertical velocity and its vertical motion will simply be a free fall. Given that the only force acting on the particle is gravity, the particle will continue to move horizontally at a constant speed until affected by another force or collision. Therefore, the question of where the particle's turning point is does not apply in a horizontal direction since it will not turn until it hits the ground or another obstacle.