greywolfboss
Answered

2) A ball is thrown straight up into the air and passes a window 3 seconds after being released. It passes the same
window on its way back down 1.5 seconds later. What was the initial velocity of the ball?

Answer :

Answer:

Hi, There! your Answer is Below!

10.29 m/s

Explanation:

a = -9.8

v = Vi - 9.8 t

h = 0 + Vi t - 4.9 t^2

-------------------------------

----------------------

at the top, t = .3 + 1.5/2 = 1.05

and v = 0 = Vi - 9.8(1.05)

so

Vi = 9.8(1.05) = 10.29 m/s

(Dont forget to mark thanks) :)

The initial velocity will be = u = 36.75 m/s

What are kinematics equations of motion ?

That describe the motion of point , bodies , and system of bodies without considering the force that cause them to move.

let that window at height h

it took 3 sec to reach that window and after crossing that window it took 1.5 second to come back to that same window

this implies , total time taken to to get to the top = 3sec + 1.5/2 = 3.75 sec

using equation of motion :

v = u + at

v= final velocity  =0(at top )

u = initial velocity =?

a = acceleration , a=g ( acceleration due to gravity) = 9.8 m/s^2

t= 3.75 sec

0 = u - 9.8(3.75)

u = 36.75 m/s

initial velocity will be = u = 36.75 m/s

learn more about kinematics equations of motion

https://brainly.com/question/13671823?referrer=searchResults

#SPJ2

Other Questions