a/ \(t=\sqrt{\dfrac{2h}{g}}=5\left(s\right)\)

b/ \(s_1=\dfrac{1}{2}gt_1^2=\dfrac{1}{2}.10.2^2=20\left(m\right)\)

c/ \(s_1'=s_1-\dfrac{1}{2}.10.1^2=15\left(m\right)\)

d/ \(s_2=h-s_3'=125-\dfrac{1}{2}.10.4^2=45\left(m\right)\)

e/ \(s_3=h-\dfrac{1}{2}.10.3^2=80\left(m\right)\)

f/ \(t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2.45}{10}}=3\left(s\right)\)

g/ \(s_4=h-s'=125-45=80\left(m\right)\)

\(t'=\sqrt{\dfrac{2h}{g}}-\sqrt{\dfrac{2h'}{g}}=1\left(s\right)\)

\(t=\dfrac{v}{g}=\dfrac{25}{9,8}=2,55\left(s\right)\)

\(h=\dfrac{1}{2}gt^2=31,9\left(m\right)\)

Như Khương Nguyễn ui bố onl lại rồi à =))) rảnh kèm con Toán Lí Hóa nhaaaa 

\(v=v_o-at\Leftrightarrow v_o=at=0,7.30=21\left(m\backslash s\right)\)

\(v^2-v_o^2=2as\) \(\Leftrightarrow s=\dfrac{v_o^2}{2a}=\dfrac{21^2}{2.0,7}=315\left(m\right)\)

\(\omega=\sqrt{\dfrac{k}{m}}\Leftrightarrow k=100\left(N\backslash m\right)\)

\(F_{dh}=kx=100.3=300\left(N\right)\)

Vậy...

a/ \(v=v_o+at\Leftrightarrow a=\dfrac{v-v_o}{t}=2\left(m\backslash s^2\right)\)

b/ \(s=\dfrac{1}{2}at^2=16\left(m\right)\)

\(v=v_o+at=2.4=8\left(m\backslash s\right)\)

c/ \(v=v_o+at\Leftrightarrow t=\dfrac{v-v_o}{a}=\dfrac{20}{2}=10\left(s\right)\)

a/ \(x_1=12t\left(km\right)\) ; \(x_2=8+4t\left(km\right)\)

b/ Lúc 6h 30p 2 người cách nhau : 

\(x_2-x_1=\left|8+4.0,5-12.0,5\right|=4\left(km\right)\)

c/ 2 người gặp nhau :

\(x_1=x_2\)

\(\Leftrightarrow12t=8+4t\Leftrightarrow t=1\left(h\right)\)

Lúc 7h 2 người gặp nhau

Vị trí cách A : 12km

2/ \(v=\dfrac{s}{t}=\dfrac{40}{1,5}=\dfrac{80}{3}\left(km\backslash h\right)=7,4\left(m\backslash s\right)\)

3/ Đổi : \(t=10p=\dfrac{1}{6}h\)

\(s=vt=60.\dfrac{1}{6}=10\left(km\right)\)