Question

what is acceleration? *
1 poin
velocity

displacement

change in velocity

change in position

what is the acceleration due to gravity on earth? *
1 point
-9.8 m/s^2

9.8 m/s^2

3.4 m/s^2

-3.4 m/s^2

a baseball is first dropped from a height of 2 meters. then, the same

• 1 point

baseball is dropped from a height of 4 meters. in which scenario will the
baseball hit the ground at a faster velocity?

your answer

Explanation:

First Question: What is acceleration?

Acceleration is defined as the rate of change of velocity. Velocity is speed with direction, displacement is change in position, and change in position is related to displacement. So the correct option is the one about change in velocity.

The acceleration due to gravity on Earth is approximately \( 9.8 \, \text{m/s}^2 \) (positive if we take the downward direction as positive, or -9.8 if upward is positive, but the standard magnitude is 9.8 m/s² for the acceleration due to gravity near the Earth's surface, and often the value is taken as \( 9.8 \, \text{m/s}^2 \) or \( -9.8 \, \text{m/s}^2 \) depending on the coordinate system. However, the magnitude is 9.8, and if we consider the direction (downward as negative in some cases, but the common value for the acceleration due to gravity's magnitude is 9.8 m/s², and the option with -9.8 is also correct in a coordinate system where upward is positive. But typically, the acceleration due to gravity is \( 9.8 \, \text{m/s}^2 \) downward, so if we take downward as positive, it's 9.8, if upward as positive, it's -9.8. However, the standard value for the acceleration due to gravity on Earth is approximately \( 9.8 \, \text{m/s}^2 \) (or \( -9.8 \, \text{m/s}^2 \) depending on sign convention). But the options include -9.8 m/s² and 9.8 m/s². The acceleration due to gravity is a vector, and if we take the upward direction as positive, then the acceleration due to gravity (which acts downward) is \( -9.8 \, \text{m/s}^2 \). However, sometimes the magnitude is referred to as 9.8 m/s². But in physics, when considering the acceleration due to gravity, the value is approximately \( 9.8 \, \text{m/s}^2 \) (or \( -9.8 \, \text{m/s}^2 \) depending on the coordinate system). But the correct options here are either -9.8 m/s² or 9.8 m/s². The standard acceleration due to gravity near the Earth's surface is \( 9.8 \, \text{m/s}^2 \) (if downward is positive) or \( -9.8 \, \text{m/s}^2 \) (if upward is positive). However, the most common value used (magnitude) is \( 9.8 \, \text{m/s}^2 \), but the option with -9.8 is also correct in a coordinate system where upward is positive. But typically, in many cases, the acceleration due to gravity is taken as \( 9.8 \, \text{m/s}^2 \) (downward), so the option with 9.8 m/s² or -9.8 m/s². But the correct answer is that the acceleration due to gravity on Earth is approximately \( 9.8 \, \text{m/s}^2 \) (or \( -9.8 \, \text{m/s}^2 \) depending on sign convention). However, the options are: -9.8 m/s², 9.8 m/s², 3.4 m/s², -3.4 m/s². So the correct options are either -9.8 m/s² or 9.8 m/s². The acceleration due to gravity is \( 9.8 \, \text{m/s}^2 \) (magnitude) acting downward, so if we take upward as positive, it's -9.8 m/s², if downward as positive, it's 9.8 m/s². Both are correct, but the standard value is \( 9.8 \, \text{m/s}^2 \) (or \( -9.8 \, \text{m/s}^2 \)).

We can use the kinematic equation \( v^2 = u^2 + 2as \), where \( u = 0 \) (dropped, so initial velocity is 0), \( a = g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), and \( s \) is the height (displacement). So \( v = \sqrt{2gh} \). Since \( g \) is constant, the velocity \( v \) is proportional to \( \sqrt{h} \). So a larger height \( h \) will result in a larger final velocity. So when dropped from 4 meters, the height is larger, so the final velocity will be faster.

Answer:

The correct option is: change in velocity (the third option among the four, with the text "change in velocity").

Second Question: What is the acceleration due to gravity on Earth?