In the formula [tex]v_y^{\prime} = a_y \Delta t[/tex], what is the value of [tex]a_y[/tex] for an object in projectile motion?

A. [tex]-9.8 \, \text{m/s}^2[/tex]
B. [tex]0 \, \text{m/s}^2[/tex]
C. [tex]\frac{1}{2} \, \text{m/s}^2[/tex]
D. [tex]1.0 \, \text{m/s}^2[/tex]



Answer :

To determine the vertical acceleration [tex]\(a_y\)[/tex] for an object in projectile motion, let's break down the components involved:

1. Understanding Projectile Motion:
- Projectile motion refers to the motion of an object that is thrown or projected into the air and is subject to only the acceleration due to gravity.

2. Vertical Acceleration [tex]\(a_y\)[/tex]:
- In the context of projectile motion, the only force acting on the object in the vertical direction (ignoring air resistance) is gravity. This leads to a constant vertical acceleration.

3. Constant Acceleration Due to Gravity:
- The acceleration due to gravity on Earth is approximately [tex]\(9.8 \, m/s^2\)[/tex].
- Since this acceleration acts downward, it is considered negative when using a coordinate system where upward is positive.

Given that the vertical acceleration [tex]\(a_y\)[/tex] for an object in projectile motion is due to gravity and acts downward, it is:

[tex]\[ a_y = -9.8 \, m/s^2 \][/tex]

Therefore, the value of [tex]\(a_y\)[/tex] for an object in projectile motion is [tex]\(-9.8 \, m/s^2\)[/tex].

So, the correct answer is:
[tex]\[ -9.8 \, m/s^2 \][/tex]