Question

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.

Answer 1

Answer:

Explanation:

Given

Dolphin leaps out an angle $\theta =36.6^{\circ}$

Horizontal component of dolphin velocity $u_x=7.87\ m/s$

Suppose u is the launch velocity of dolphin

therefore $u\cos \theta =u_x---1$

and vertical velocity $u_y=u\sin \theta ----2$

divide 1 and 2 we get

$\tan \theta =\frac{u_y}{u_x}$

$u_y=u_x\tan \theta$

$u_y=7.87\cdot \tan (36.6)$

$u_y=5.84\ m/s$