a ball is shot from the ground straight up into the air with initial velocity of 41 ft/sec. assuming that the air resistance can be ignored, how high does it go?



Answer :

PNaik

A ball is shot from the ground straight up into the air with initial velocity of 41 ft/sec. It takes 1.28 sec to reach its  its maximum height of 26.27feet.

a ball is shot straight up into the air with initial velocity of 41ft/sec

S = [tex]S_{0} +V_{t} - 16t^{2}[/tex]

height [tex]S_{0} = 0[/tex]

velocity [tex]V_{0}[/tex] = 41ft/sec

S= 0+41t- 16[tex]t^{2}[/tex]

S = 41t -16[tex]t^{2}[/tex]

instantaneous velocity v

V = ds/dt = [tex]V_{0}[/tex]- 32t

V = 41-32t

32t = 41

t =  41/32 = 1.28 sec.

S = 41t-16[tex]t^{2}[/tex]

  = 41 ( 1.28)- 16 [tex](1.28)^{2}[/tex]

 =  52.48 - 26.21

 = 26.27

What is velocity?

The pace at which an object's position changes in relation to a frame of reference and time is what is meant by velocity. Although it may appear sophisticated, velocity is just the act of moving quickly in one direction. Since it is a vector quantity, the definition of velocity requires both magnitude (speed) and direction. Its SI equivalent is the meter per second (ms-1). A body is considered to be accelerating if the magnitude or direction of its velocity changes.

To learn more about velocity, refer;

https://brainly.com/question/18084516

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