a ball is thrown straight up into the air with an initial speed of 3.1 m/s. a. After 0.24 seconds what is the ball's velocity and b. what is it's acceleration?

Respuesta :

a. The ball's velocity after 0.24 s is 0.75 m/s

b. The acceleration of the ball is given by the acceleration due to gravity

a. The ball's velocity can be calculated with the following equation:

[tex] v_{f} = v_{0} - gt [/tex]

Where:

[tex] v_{f} [/tex]: is the final speed =?

[tex]v_{0}[/tex]: is the initial speed = 3.1 m/s

g: is the acceleration due to gravity = 9.81 m/s²

t: is the time = 0.24 s

The minus sign is because the acceleration is in the opposite direction (downward) of the motion of the ball (upward).

The final speed is:

[tex] v_{f} = v_{0} - gt = 3.1 m/s - 9.81 m/s^{2}*0.24 s = 0.75 m/s [/tex]

Hence, the ball's velocity after 0.24 s is 0.75 m/s.

b. The acceleration of the ball is given by the acceleration due to gravity because the ball is thrown straight up (the motion of the ball is in the y-direction). The velocity of the ball in the x-direction is zero so the acceleration in the same direction is also zero.  

You can see another example of velocity here: https://brainly.com/question/13388351?referrer=searchResults

I hope it helps you!