Đáp án:

\(\begin{array}{l}
a.{v_{\max }} = 10\sqrt 2 m/s\\
b.h = 5m\\
c.h = \frac{{10}}{3}m\\
v = 11,547m/s
\end{array}\)

Giải thích các bước giải:

a.

Bảo toàn cơ năng:

\(\begin{array}{l}
{W_{t\max }} = {W_{d\max }} \Leftrightarrow mg{h_{\max }} = \frac{1}{2}mv_{\max }^2\\
 \Rightarrow 10.10 = \frac{1}{2}v_{\max }^2 \Rightarrow {v_{\max }} = 10\sqrt 2 m/s
\end{array}\)

b.

Ta có:

\(\begin{array}{l}
{W_t} = {W_d} = \frac{W}{2} = \frac{{{W_{t\max }}}}{2} \Rightarrow mgh = \frac{{mg{h_{\max }}}}{2}\\
 \Rightarrow h = \frac{{{h_{\max }}}}{2} = \frac{{10}}{2} = 5m
\end{array}\)

c.

Ta có:

\(\begin{array}{l}
{W_d} = 2{W_t} \Rightarrow {W_t} = \frac{W}{3} = \frac{{{W_{t\max }}}}{3} \Rightarrow mgh = \frac{{mg{h_{\max }}}}{3}\\
 \Rightarrow h = \frac{{{h_{\max }}}}{3} = \frac{{10}}{3}m\\
{W_d} = 2{W_t} \Rightarrow \frac{1}{2}m{v^2} = 2mgh\\
 \Rightarrow \frac{1}{2}.{v^2} = 2.10.\frac{{10}}{3} \Rightarrow v = 11,547m/s
\end{array}\)

Cơ năng vật:

$W_A=Wt_A=mgh_A=20.10.10=2000J$

a. Vận tốc chạm đất:

$W_A=W_B=Wđ_B=\frac{1}{2}mv_B^2$

$\Leftrightarrow 2000=\frac{1}{2}.20.v_B^2$

$\Rightarrow v_B=10\sqrt{2} m/s$

b. Độ cao vật:

$W_A=W_C=2Wt_C=2mgh_C$

$\Leftrightarrow 2000=2.20.10.h_C$

$\Rightarrow h_C=5m$

c. Độ cao vật:

$W_A=W_D=3Wt_D=3mgh_D$

$\Leftrightarrow 2000=3.20.10.h_D$

$\Rightarrow h_D=\frac{10}{3}m$

Vận tốc vật:

$\frac{1}{2}mv_D^2=W-Wt_D=2000-20.10.\frac{10}{3}$

$\Leftrightarrow \frac{1}{2}.20.v_D^2=\frac{4000}{3}$

$\Rightarrow v_D≈11,54m/s$