Đáp án:

\(\begin{array}{l}
a.{v_{\max }} = 20m/s\\
b.h = 10m\\
c.v' = 10m/s
\end{array}\)

Giải thích các bước giải:

 a.

Bảo toàn cơ năng:

\(\begin{array}{l}
W = {W_{d\max }} \to mg{h_{\max }} = \frac{1}{2}mv_{\max }^2\\
 \to 10.20 = \frac{1}{2}v_{\max }^2 \Rightarrow {v_{\max }} = 20m/s
\end{array}\)

b.

Ta có:

\(\begin{array}{l}
{W_t} = {W_d} = \frac{W}{2} \Rightarrow mgh = \frac{{mg{h_{\max }}}}{2}\\
 \Rightarrow h = \frac{{{h_{\max }}}}{2} = \frac{{20}}{2} = 10m
\end{array}\)

c.

Ta có:

\(\begin{array}{l}
{W_t}' = 3{W_d}' \Rightarrow {W_d}' = \frac{W}{4} = \frac{{{W_{d\max }}}}{4} \to \frac{1}{2}mv{'^2} = \frac{{\frac{1}{2}mv_{\max }^2}}{4}\\
 \Rightarrow v{'^2} = \frac{{v_{\max }^2}}{4} \Rightarrow v' = \frac{{{v_{\max }}}}{2} = \frac{{20}}{2} = 10m/s
\end{array}\)

a. Cơ năng vật:

$W_A=Wt_A=mgh_A=10.20.m=200m(J)$

Vận tốc khi chạm đất:

$W_A=W_B=Wđ_B=\frac{1}{2}mv_B^2$

$\Leftrightarrow 200=\frac{1}{2}v_B^2$

$\Rightarrow v_B=20(m/s)$

b. $W_A=W_C=2Wt_C=2mgh_C$

$\Leftrightarrow 200=2.10.h_C$

$\Rightarrow h_C=10(m)$

c. $W_A=W_C=4Wd_C=4.\frac{1}{2}mv_C^2$

$\Leftrightarrow 200=2.v_C^2$

$\Rightarrow v_C=10(m/s)$