Question

A diver drops from a 10 meter high cliff. At what speed does he enter the
water, and how long is he in the air?

Answer 1

just use s=ut+1/2at^2
u=0;a=10;s=10

Answer 2

use equation of continuity
\[\rho gh+1/2\rho v ^{2}+P=const.\]
rho=density of water P=pressure ,v = speed h =relative height

Answer 3

I like using potential energy = kinetic energy
\[mgh=\frac{ 1 }{ 2 }mv ^{2}\]
\[gh=\frac{ 1 }{ 2 }v ^{2}\]
\[v ^{2}=2gh\]
So speed =
\[v=\sqrt{2gh}\]
Using speed to calculate time in the air:
\[h=vt\]
\[t=\frac{ h }{ v }\]

Answer 4

if the diver drops, shouldn't the diver be following the typical accelaration of y=yi+vi*t+1/2at^2. Since he drops, he accelerates at 9.81 m/s^2 and starts at a velocity of 0. Therefore y=10-1/2*a*t^2. Since y=10, 10=1/2*9.81*t^2 so (20/9.81)=t^2 and sqrt(20/9.81)=t.

The overall algebraic equation should then be sqrt((2*y)/a)=t considering that the diver starts from rest and a=-9.81m/s^2

Answer 5

See above:
a=-9.81 m/s^2
v(initial)=0;
y(initial)=10

Overall equation
\[y=yi+vi*t+1/2*a*t ^{2}\]

Velocity Equation

\[ \sqrt{(2*y)/a}=t \]