# "A ball is thrown up from the rooftop"

• Feb 13th 2010, 07:25 PM
s3a
"A ball is thrown up from the rooftop"
"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!

• Feb 13th 2010, 11:09 PM
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!

Using the equation
$\displaystyle s=ut+\tfrac12at^2$
and taking $\displaystyle t = 0$ when the first ball is thrown, the height $\displaystyle h_1$ of the first ball above the roof at time $\displaystyle t$ seconds is given by:
$\displaystyle h_1=10t-4.9t^2$, taking $\displaystyle g = 9.8$
The height $\displaystyle h_2$ of the second ball above the roof at time $\displaystyle t$ is:
$\displaystyle h_2 = -4.9(t-1.84)^2$
They meet when $\displaystyle h_1= h_2$; i.e. when:
$\displaystyle 10t-4.9t^2=-4.9(t-1.84)^2$
$\displaystyle =-4.9t^2+18.032t-16.58944$
$\displaystyle \Rightarrow 8.032t = 16.58944$

$\displaystyle \Rightarrow t = 2.0654$
So, as you say, they meet after $\displaystyle 2.0654$ seconds.

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

Using $\displaystyle v = u+at$, the speed of the first ball is:
$\displaystyle 10-9.8\times2.0654=-10.24$
i.e. $\displaystyle 10.24$ m/sec.

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

i.e. $\displaystyle 2.209$ m/sec