Question

A ball 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.

Answer 1

`h=53.09m` (2)

`v_(min)>5.05m/s`

`v_(max)<10.4m/s`

Step-by-step explanation:

a)Kinematics equation for the first ball:

`v(t)=v_(o)-g*t`

`y(t)=y_(o)+v_(o)t-1/2*g*t^(2)`

`y_(o)=h` initial position is the building height

`v_(o)=8.9m/s`

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

`0=h+v_(o)t_(1)-1/2*g*t_(1)^(2)`

`h=1/2*g*t_(1)^(2)-v_(o)t_(1)` (1)

Kinematics equation for the second ball:

`v(t)=v_(o)-g*t`

`y(t)=y_(o)+v_(o)t-1/2*g*t^(2)`

`y_(o)=h` initial position is the building height

`v_(o)=0` the ball is dropped

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

`0=h-1/2*g*t_(2)^(2)`

`h=1/2*g*t_(2)^(2)` (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):

`1/2*g*t_(1)^(2)-v_(o)t_(1)=1/2*g*t_(2)^(2)=1/2*g*(t_(1)-1.03)^(2)`

`g*t_(1)^(2)-2v_(o)t_(1)=g*(t_(1)^(2)-2.06*t_(1)+1.06)`

`g*t_(1)^(2)-2v_(o)t_(1)=g*(t_(1)^(2)-2.06*t_(1)+1.06)`

`-2v_(o)t_(1)=g*(-2.06*t_(1)+1.06)`

`2.06*gt_(1)-2v_(o)t_(1)=g*1.06`

`t_(1)=g*1.06/(2.06*g-2v_(o))`

vo=8.9m/s

`t_(1)=9.81*1.06/(2.06*9.81-2*8.9)=4.32s`

t2=t1-1.03 (3)

t2=3.29sg

`h=1/2*g*t_(2)^(2)=1/2*9.81*3.29^(2)=53.09m` (2)

b)
`t_(1)=g*1.06/(2.06*g-2v_(o))`

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

`1.03>9.81*1.06/(2.06*g-2v_(o))`

`1.03*(2.06*9.81-2v_(o))<9.81*1.06`

`20.8-2.06v_(o)<10.4`

`(20.8-10.4)/2.06<v_(o)`

`v_(min)>5.05m/s`

limit case: t1>0:

`g*1.06/(2.06*g-2v_(o))>0`

`2.06*g-2v_(o)>0`

`v_(o)<1.06*9.81`

`v_(max)<10.4m/s`