Aball is thrown straight up from the edge of the roof of a building. a second ball is dropped from the roof a time of 1.03 s later. you may ignore air resistance. a-if the initial speed of the first ball is v0 = 8.90 m/s what must the height h of the building be for both balls to reach the ground at the same time?
b-if v0 is greater than some value vmax, a value of h does not exist that allows both balls to hit the ground at the same time. solve for vmax
c-if v0 is less than some value vmin, a value of h does not exist that allows both balls to hit the ground at the same time. solve for vmin.

        (2)

Explanation:

a)Kinematics equation for the first ball:

      initial position is the building height

The ball reaches the ground, y=0, at t=t1:

          (1)

Kinematics equation for the second ball:

      initial position is the building height

      the ball is dropped

The ball reaches the ground, y=0, at t=t2:

        (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03              (3)

We solve the equations (1) (2) (3):

vo=8.9m/s

t2=t1-1.03              (3)

t2=3.29sg

        (2)

b)

t1 must :   t1>1.03  and t1>0

limit case: t1>1.03:

limit case: t1>0:

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation

friction

explanation:

they us to move forward and backwards