We appreciate your visit to MCQ 9 When a ball with projectile motion is rising up it A decelerates B accelerates C rises up with constant acceleration D acceleration becomes. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Below is the detailed explanation for each question:
1. For the rising projectile, while it moves upward the only force acting (gravity) is downward. This means that the upward velocity is reduced over time, or in other words, the projectile decelerates. Hence, the answer is:
[tex]$$\textbf{A. decelerates.}$$[/tex]
2. The equations of motion (often called the kinematic equations) are derived from the laws of mechanics and are applicable to bodies undergoing constant acceleration. These equations can be used to describe motion along a straight line, on a curve, or even along a circular path. Therefore, we use them for all types of motion. The answer is:
[tex]$$\textbf{D. all types of motion.}$$[/tex]
3. The acceleration due to free fall (commonly denoted by [tex]$g$[/tex]) is given by the expression
[tex]$$ g = \frac{GM}{R^2} $$[/tex]
where [tex]$R$[/tex] is the distance from the center of the Earth. This relation shows that free-fall acceleration depends on the distance from the center of the Earth. So, the correct answer is:
[tex]$$\textbf{C. distance from center of Earth.}$$[/tex]
4. When the initial velocity is zero and the object experiences a constant acceleration of [tex]$9.8\ \text{m/s}^2$[/tex], the distance [tex]$s$[/tex] traveled in time [tex]$t$[/tex] is given by the equation
[tex]$$ s = \frac{1}{2}at^2. $$[/tex]
Substituting [tex]$a = 9.8\ \text{m/s}^2$[/tex], we have
[tex]$$ s = \frac{1}{2} \times 9.8 \times t^2 = 4.9t^2. $$[/tex]
Thus the distance covered is [tex]$4.9t^2$[/tex], which corresponds to:
[tex]$$\textbf{D. }4.9t^2.$$[/tex]
5. As the ball falls toward the ground, gravity causes it to speed up. In other words, its velocity increases with time. Therefore, the answer is:
[tex]$$\textbf{A. increases.}$$[/tex]
In summary, the answers are:
MCQ 9: A
MCQ 10: D
MCQ 11: C
MCQ 12: D
MCQ 13: A
1. For the rising projectile, while it moves upward the only force acting (gravity) is downward. This means that the upward velocity is reduced over time, or in other words, the projectile decelerates. Hence, the answer is:
[tex]$$\textbf{A. decelerates.}$$[/tex]
2. The equations of motion (often called the kinematic equations) are derived from the laws of mechanics and are applicable to bodies undergoing constant acceleration. These equations can be used to describe motion along a straight line, on a curve, or even along a circular path. Therefore, we use them for all types of motion. The answer is:
[tex]$$\textbf{D. all types of motion.}$$[/tex]
3. The acceleration due to free fall (commonly denoted by [tex]$g$[/tex]) is given by the expression
[tex]$$ g = \frac{GM}{R^2} $$[/tex]
where [tex]$R$[/tex] is the distance from the center of the Earth. This relation shows that free-fall acceleration depends on the distance from the center of the Earth. So, the correct answer is:
[tex]$$\textbf{C. distance from center of Earth.}$$[/tex]
4. When the initial velocity is zero and the object experiences a constant acceleration of [tex]$9.8\ \text{m/s}^2$[/tex], the distance [tex]$s$[/tex] traveled in time [tex]$t$[/tex] is given by the equation
[tex]$$ s = \frac{1}{2}at^2. $$[/tex]
Substituting [tex]$a = 9.8\ \text{m/s}^2$[/tex], we have
[tex]$$ s = \frac{1}{2} \times 9.8 \times t^2 = 4.9t^2. $$[/tex]
Thus the distance covered is [tex]$4.9t^2$[/tex], which corresponds to:
[tex]$$\textbf{D. }4.9t^2.$$[/tex]
5. As the ball falls toward the ground, gravity causes it to speed up. In other words, its velocity increases with time. Therefore, the answer is:
[tex]$$\textbf{A. increases.}$$[/tex]
In summary, the answers are:
MCQ 9: A
MCQ 10: D
MCQ 11: C
MCQ 12: D
MCQ 13: A
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