hey everyone,
im not here to just get answers because im lazy, im here because i got a 10 page notebook and ive done 90 pct of the questions and just a few is left. This question really confuses me, i just need help to solve it so i can be done
thnx if u help me(im new here)

A student designs a special container as a part of an egg drop experiment. she believes the container can withstand a fall as long as the speed of the container does not exceed 80 ft/s. shes uses this equation
v=sqrt(v0)2+2ad (the whole thing is square root)
to model the velocity v, in feet per second, as a function of constant acceleration, a, in feet per second squared and the drop distance d in feet. the acceleration due to gravity is 32ft/s2

a) will the egg break if the student drops the egg from shoulder height(5 ft) off a building 80 ft high?

i did so far
v=sqrt(v0)2+2(32)(85)

b) what is the maximum height the egg can be dropped from?

no help guys?

You are on the right track...now if the egg is "dropped" what is $v_0$?

Hello, EASYPEAAZY!

A student designs a special container as a part of an egg-drop experiment.
She believes the container can withstand a fall as long as the speed of the container does not exceed 80 ft/s.
She uses this equation: $v\:=\:\sqrt{v_o^2 +2ad}$ to model the velocity $v$ in ft/sec,
as a function of constant acceleration, $a = 32 \text{ft/sec}^2$ and the drop distance $d$ in feet.

a) Will the egg break if the student drops the egg from shoulder height (5 ft) off a building 80 ft high?

I did so far: $v\:=\:\sqrt{v_o^2+2(32)(85)}$

Note that the egg is dropped.
That is: . $v_o \,=\,0$

b) What is the maximum height the egg can be dropped from?

We have: . $\sqrt{0^2+64d} \:\le\:80$

Square: n . . . . . $64d \:\le\:6400$

Therefore: . . . . . . $d \:\le\:100$

ah i see velocity is 0, i just remembered that thnx

a) so i solved it and got v=sqrt(64)(85).....-> v=sqrt(5440)----->v=73.76 ft/s...so it will not break (thnx mark and soroban)

^ tell me if its right

b) i did 80=sqrt(0^2+64d)---> 80^2=sqrt(64d)^2--> 6400=64d--->d=100... 100 is the maximum height that it will not break(its a lil different than what you did soroban)

tell me if those are right please(thnx to the 2 posters above me for giving me help)

$v_f=8\sqrt{85}\text{ }\frac{\text{ft}}{\text{s}}$