"A ball is thrown up from the rooftop"

Hello s3a

Quote:

Originally Posted by s3a

"A ball is thrown up from the rooftop with an initial speed of http://gauss.vaniercollege.qc.ca/web...fe94d858a1.png m/s. 1.84 s later, another ball is dropped from the rooftop.

a) Assuming that neither has landed, where and when do they meet?
m from the rooftop
at seconds from the moment the first ball was thrown.

b) What are their speeds when they meet?
The first ball : m/s
The second ball : m/s"

All my answers are wrong, except t = 2.065. I would appreciate it a lot if someone could help me do this question!

Thanks in advance!

Using the equation
$s=ut+\tfrac12at^2$
and taking $t = 0$ when the first ball is thrown, the height $h_1$ of the first ball above the roof at time $t$ seconds is given by:
$h_1=10t-4.9t^2$ , taking $g = 9.8$

The height $h_2$ of the second ball above the roof at time $t$ is:
$h_2 = -4.9(t-1.84)^2$

They meet when $h_1= h_2$ ; i.e. when:
$10t-4.9t^2=-4.9(t-1.84)^2$
$=-4.9t^2+18.032t-16.58944$
$\Rightarrow 8.032t = 16.58944$

$\Rightarrow t = 2.0654$

So, as you say, they meet after $2.0654$ seconds.

When $t = 2.0654$ ,
$h_2 = -4.9(2.065-1.84)^2=-0.249$
So the meet $0.249$ m below the level of the roof.

Using $v = u+at$ , the speed of the first ball is:

$10-9.8\times2.0654=-10.24$
i.e. $10.24$ m/sec.

And the speed of the second $= 0-9.8\times(2.0654-1.84) =-2.209$

i.e. $2.209$ m/sec

Grandad