giải
a) có: \(W=\) \(\text{ W}\text{d}_{max}=\text{ W}\text{t}_{max}\)
\(\Rightarrow\frac{1}{2}.m.v^2=m.g.h_{max}\)
\(\Rightarrow\frac{1}{2}.8^2=1.10.h_{max}\)
\(\Rightarrow h_{max}=3,2m\)
b) \(\text{ W}=\text{ W}d_{max}=\text{ W}\text{d}+\text{ W}\text{t}\)
\(\frac{1}{2}.m.v^2=2\text{ W}\text{t}\text{ }\)
\(\frac{1}{2}.8^2=2.10.h'\Rightarrow h'=1,6m\)