# A problem about free fall

• Jul 1st 2007, 05:46 AM
A problem about free fall
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50.0m below its starting point 6.00 seconds after it leaves the thrower's hand. Air resistance may be ignored.
a.) what is the initial speed of the egg? (answer: 21.2 m/s)

b.)How high does it rise above its starting point?(answer: 22.6m)

im stuck with these two questions and i got 58m/s in the 1st part ...im reallyyy stuck
• Jul 1st 2007, 06:47 AM
CaptainBlack
Quote:

Originally Posted by ^_^Engineer_Adam^_^
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50.0m below its starting point 6.00 seconds after it leaves the thrower's hand. Air resistance may be ignored.
a.) what is the initial speed of the egg? (answer: 21.2 m/s)

b.)How high does it rise above its starting point?(answer: 22.6m)

im stuck with these two questions and i got 58m/s in the 1st part ...im reallyyy stuck

You need the equation of motion which if we take $h$ positive
upward and the origin as the point from which it is threown we have:

$h''=-g$

so:

$h'=-gt+v_0$

and :

$
h=-gt^2/2+v_0t
$

We are told that $h=-50$ at $t=6$ so:

$
-50 = -18 g +6 v_0
$

so:

$v_0=\frac{18g-50}{6}$

It reaches its maximum height when $h'=0$ which is when:

$t=v_0/g$

RonL
• Jul 1st 2007, 08:07 AM
Soroban

Quote:

An egg is thrown nearly vertically upward from a point near the cornice of a tall building.
It just misses the cornice on the way down and passes a point 50.0m below its starting point
6.00 seconds after it leaves the thrower's hand.

a) What is the initial speed of the egg? (answer: 21.2 m/s) ?

b) How high does it rise above its starting point? (answer: 22.6m)

Are you familiar with the free-fall formula? . $y \;=\;h_o + v_ot - 4.9t^2$
. . where $h_o$ = initial height, $v_o$ = initial velocity,
. . and $y$ = height of object at time $t$.

Let launch point be at $h_o=0$.
. . Then we have: . $y \:=\:v_ot - 4.9t^2$

(a) We are told that $y = -50$ when $t = 6$.
. . So we have: . $-50 \;=\;v_o(6) - 4.9(6^2)\quad\Rightarrow\quad v_o \;=\;\frac{126.4}{6} \;=\;21.0666...$

Therefore, the initial speed is: . $v_o \;\approx\;\boxed{21.1\text{ m/s}}$

(b) To maximize height, equate the derivative to zero.

We have: . $y \;=\;21.1t - 4.9t^2$
Then: . $y' \;=\; 21.1 - 9.8t \;=\;0\quad\Rightarrow\quad t \;=\;\frac{21.1}{9.8} \;\approx\;2.15$ seconds

Therefore: . $y \;=\;\frac{316}{15}(2.15) - 4.9(2.15)^2 \;\approx\;\boxed{22.6\text{ feet}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you aren't allowed to use Calculus, we can still solve it.

The graph of: . $y \;=\;21.1t - 4.9t^2$ is a down-opening parabola.
. . Its maximum height occurs at its vertex.

The vertex is given by: . $t \:=\:\frac{-b}{2a}$

We have: . $a = -4.9,\;b = 21.1$

. . Hence: . $t \;=\;\frac{-21.1}{2(-4.9)} \;\approx\;2.15$ seconds . . . see?

• Jul 1st 2007, 08:19 AM