In the formula [tex]\( v_y = a_y \Delta t \)[/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 :

Certainly! Let's analyze the situation.

In projectile motion, an object is influenced by the force of gravity. Gravity plays a crucial role in determining the motion of the object, particularly in the vertical (y) direction. The standard acceleration due to gravity on the surface of the Earth is approximately [tex]\( -9.8 \, m/s^2 \)[/tex]. This negative sign indicates that gravity acts downward.

Given this understanding, let's evaluate the options provided for [tex]\( a_y \)[/tex]:

1. [tex]\( -9.8 \, m/s^2 \)[/tex]
2. [tex]\( 0 \, m/s^2 \)[/tex]
3. [tex]\( \frac{1}{2} \, m/s^2 \)[/tex]
4. [tex]\( 1.0 \, m/s^2 \)[/tex]

In projectile motion, the only vertical acceleration acting on the object is due to gravity, which we know is [tex]\( -9.8 \, m/s^2 \)[/tex]. All the other values provided do not represent the acceleration due to gravity and are therefore incorrect.

Thus, for the vertical acceleration [tex]\( a_y \)[/tex] in the context of projectile motion, the correct value is:
[tex]\[ a_y = -9.8 \, m/s^2 \][/tex]

Therefore, the correct answer is:

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

Hence, in the formula [tex]\( v_y = a_y \Delta t \)[/tex], the value of [tex]\( a_y \)[/tex] for an object in projectile motion is:

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