1. ## 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

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 $\displaystyle h$ positive
upward and the origin as the point from which it is threown we have:

$\displaystyle h''=-g$

so:

$\displaystyle h'=-gt+v_0$

and :

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

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

$\displaystyle -50 = -18 g +6 v_0$

so:

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

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

$\displaystyle t=v_0/g$

RonL

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? . $\displaystyle y \;=\;h_o + v_ot - 4.9t^2$
. . where $\displaystyle h_o$ = initial height, $\displaystyle v_o$ = initial velocity,
. . and $\displaystyle y$ = height of object at time $\displaystyle t$.

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

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

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

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

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

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

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

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

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

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

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

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

4. thank you!