Đáp án:
\(\begin{array}{l}
a.\\
{h_{\max }} = 180m\\
b.\\
v = 20\sqrt 5 m/s\\
c.\\
v' = 30m/s\\
d.\\
h = 60m\\
2.\\
a.\\
{v_{\max }} = 100m/s\\
b.\\
s = 20m\\
c.\\
h' = 375m\\
d.\\
h = 250m\\
e.\\
v' = 40\sqrt 5 m/s
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{W_{d\max }} = {W_{t\max }}\\
\frac{1}{2}mv_{\max }^2 = mg{h_{\max }}\\
\frac{1}{2}{.60^2} = 10{h_{\max }}\\
{h_{\max }} = 180m\\
b.\\
W = {W_{t\max }}\\
mgh + \frac{1}{2}m{v^2} = mg{h_{\max }}\\
10.80 + \frac{1}{{2.}}{v^2} = 10.180\\
v = 20\sqrt 5 m/s\\
c.\\
W' = {W_d}' + {W_t}' = {W_d}' + 3{W_d}' = 4{W_d}' = 4\frac{1}{2}mv{'^2}\\
{W_{t\max }} = W'\\
mg{h_{\max }} = 2mv{'^2}\\
10.180 = 2.v{'^2}\\
v' = 30m/s\\
d.\\
W'' = {W_d}'' + {W_t}'' = {W_t}' + 2{W_t}' = 3{W_t}' = 3mgh'\\
{W_{t\max }} = W''\\
mg{h_{\max }} = 3mgh\\
180 = 3h\\
h = 60m\\
2.\\
a.\\
{W_{d\max }} = {W_{t\max }}\\
\frac{1}{2}mv_{\max }^2 = mg{h_{\max }}\\
\frac{1}{2}.{v_{\max }}^2 = 10.500\\
{v_{\max }} = 100m/s\\
b.\\
W = {W_{t\max }}\\
mgh + \frac{1}{2}m{v^2} = mg{h_{\max }}\\
10.h + \frac{1}{{2.}}{20^2} = 10.500\\
h = 480m\\
s = {h_{\max }} - h = 500 - 480 = 20m\\
c.\\
W' = {W_{t\max }}\\
mgh' + \frac{1}{2}mv{'^2} = mg{h_{\max }}\\
10.h' + \frac{1}{{2.}}{50^2} = 10.500\\
h' = 375m\\
d.\\
W = {W_t} + {W_d} = 2{W_t}''\\
{W_{t\max }} = W\\
mg{h_{\max }} = 2mgh\\
500 = 2.h\\
h = 250m\\
e.\\
W' = {W_t}' + {W_d}' = \frac{1}{4}{W_d}' + {W_d}' = \frac{5}{4}{W_d}'\\
{W_{t\max }} = \frac{5}{4}{W_d}'\\
mg{h_{\max }} = \frac{5}{4}.\frac{1}{2}mv{'^2}\\
10.500 = \frac{5}{8}.v{'^2}\\
v' = 40\sqrt 5 m/s
\end{array}\)
