Ciało spada swobodnie z wysokości 5 m .Jaką szybkość osiągnie w momencie uderzania o ziemie ?
mgh=mv^2/2
gh=v^2/2
2gh^1/2=v
2*5*10^1/2=10m/s
s = 5m
a = 10m/s^2
t = ?
v = ?
s = 1/2 * a * t^2 / *2
2s = a * t^2 / /a
2s / a = t^2 / pierwiastek
pierwistek z ( 2s/a) = t
t = pierwiastek ( 2 * 5m / 10 m/s^2 )
t = pierwiastek (1 s^2)
t = 1s
V = a * t
V= 10 m/s^2 * 1s
V = 10m/s
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
mgh=mv^2/2
gh=v^2/2
2gh^1/2=v
2*5*10^1/2=10m/s
s = 5m
a = 10m/s^2
t = ?
v = ?
s = 1/2 * a * t^2 / *2
2s = a * t^2 / /a
2s / a = t^2 / pierwiastek
pierwistek z ( 2s/a) = t
t = pierwiastek ( 2 * 5m / 10 m/s^2 )
t = pierwiastek (1 s^2)
t = 1s
V = a * t
V= 10 m/s^2 * 1s
V = 10m/s