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A dolphin leaps out of the water at an angle of 36.6° above the horizontal. The horizontal component of the dolphin's velocity is 7.87 m/s. Calculate the magnitude of the vertical component of the velocity.

Respuesta :

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]