Answer:
Explanation:
Given
Dolphin leaps out an angle [tex]\theta =36.6^{\circ}[/tex]
Horizontal component of dolphin velocity [tex]u_x=7.87\ m/s[/tex]
Suppose u is the launch velocity of dolphin
therefore [tex]u\cos \theta =u_x---1[/tex]
and vertical velocity [tex]u_y=u\sin \theta ----2[/tex]
divide 1 and 2 we get
[tex]\tan \theta =\frac{u_y}{u_x}[/tex]
[tex]u_y=u_x\tan \theta[/tex]
[tex]u_y=7.87\cdot \tan (36.6)[/tex]
[tex]u_y=5.84\ m/s[/tex]
