Question

1.) what happens as an object free falls? a. velocity increases b. acceleration increases c. both of these d. none of these 2.) what is the acceleration of a car that maintains a constant velocity of 100 m/s for 10 s? a. 0 m/s² b. 10 m/s² c. 100 m/s² d. 1000 m/s² 3.) a ball is tossed vertically into the air. when the ball reaches its highest point, what is the acceleration of the ball? a. 0 m/s² b. 10 m/s² downward c. 15 m/s² - downward d. 5 m/s² - downward 4.) george and jerry are on a high bridge above a river. george drops a baseball off of the bridge at the time that jerry throws a baseball downward off of the bridge. ignoring air resistance, what is the acceleration of the balls are released? a. greater for the dropped baseball b. greater for the thrown baseball c. the same for both baseballs d. depends on how height of the bridge 5.) what happens to a stone each second that it free falls? a. travels the same distance as in the second before b. has the same average speed c. has a larger distance than the second before d. has the same instantaneous speed

Explanation:

Step1: Analyze free - fall motion

In free - fall (ignoring air resistance), the acceleration due to gravity \(g\approx10m/s^{2}\) is constant near the Earth's surface. Velocity \(v = v_0+gt\) (where \(v_0\) is the initial velocity and \(t\) is time), so as time \(t\) increases, velocity increases.

Step2: Analyze constant velocity case

Acceleration \(a=\frac{\Delta v}{\Delta t}\). If velocity \(v\) is constant (\(\Delta v = 0\)), then \(a = 0\).

Step3: Analyze ball at highest point

When a ball is tossed vertically, at the highest point, the velocity is \(0\) for an instant, but the acceleration due to gravity is still \(g\approx10m/s^{2}\) downward.

Step4: Analyze dropped and thrown balls

The acceleration of any object in free - fall near the Earth's surface (ignoring air resistance) is the acceleration due to gravity \(g\), regardless of how it is launched.

Step5: Analyze free - falling stone

The distance traveled by a free - falling object in time \(t\) is \(d=v_0t+\frac{1}{2}gt^{2}\). In each successive second, the distance traveled increases because of the \(\frac{1}{2}gt^{2}\) term.

Answer:

1. A. velocity increases
2. A. \(0m/s^{2}\)
3. B. \(10m/s^{2}\) downward
4. C. the same for both baseballs
5. C. has a larger distance than the second before