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A ball is hit straight up with an initial speed of 28 meters per second. What is the speed of the ball 2.2 seconds after it is hit? [Neglect friction.]

Respuesta :

taskmasters
By neglecting friction, the final velocity of the ball thrown upward with an initial speed of v₀ can be calculated through the equation,
                                          vf = v₀ - gt
where vf is final velocity, g is acceleration due to gravity and t is time. Substituting,
                                         vf = 28 m/s - (9.8 m/s²)(2.2. s)
                                           vf = 6.44 m/s
Thus, the final velocity of the ball is approximately 6.44 m/s. 

Answer:

The velocity after 2.2sec is [tex]v=6.44 \rm m/s[/tex]

Explanation:

Given information:

Initial speed of ball is [tex]u=28\rm m/s[/tex]

The ball is moving straight upward. So, acceleration due to gravity will act in opposite direction to the motion.

So, the acceleration of the ball will be,

[tex]a=-g\\a=-9.8\rm m/s^2[/tex]

It is required to find the velocity [tex]v=?[/tex]

Time [tex]t=2.2 \rm sec[/tex]

By using first equation of motion

[tex]v=u+at[/tex]

On substituting,

[tex]v=28\rm m/s -(9.8\rm m/s^{2} )\times2.2\rm s[/tex]

[tex]v=6.44\rm m/s[/tex]

Hence, the velocity after 2.2 sec is [tex]v=6.44\rm m/s[/tex] .

For more details, refer the link:

https://brainly.com/question/13704950?referrer=searchResults

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