Answer:
6.6 m/s
Explanation:
We are given that
Length,l=14 m
[tex]\theta=45^{\circ}[/tex]
[tex]\theta'=30^{\circ}[/tex]
We have to find the Tarzan's speed before he reaches Jane.
Difference in height,h=Final height-Initial height=l(cos 30-cos 45)
Substitute the values
h=14(cos30-cos45)=2.22 m
Speed of Tarzan is given by
[tex]v=\sqrt{2gh}[/tex]
Where [tex]g=9.8m/s^2[/tex]
Substitute the values
[tex]v=\sqrt{2\times 9.8\times 2.22}=6.6 m/s[/tex]
Hence, Tarzan's speed just before he reaches Jane=6.6 m/s