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

Step-by-step 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 =(u_y)/(u_x)`

`u_y=u_x\tan \theta`

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

`u_y=5.84\ m/s`