# If the dolphin is moving horizontally when it goes through the hoop, how high above the water is the center of the hoop?A dolphin jumps with an initial velocity of 12.0 m/s at an angle of 39.3...

If the dolphin is moving horizontally when it goes through the hoop, how high above the water is the center of the hoop?

A dolphin jumps with an initial velocity of 12.0 m/s at an angle of 39.3 above the horizontal. The dolphin passes through the center of a hoop before returning to the water.

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The initial velocity is 12.0 m/s with an angle of 39.3 degrees above the horizontal. This means that the initial velocity in the vertical direction is:

`v_y=12.0sin(39.3)=7.6`

Since the dolphin is moving horizontally when it goes through the hoop, it has vertical velocity of 0 at that point. The vertical velocity equation is

`v=-g t + v_y` sub in values

`0=-9.8t+7.6` solve for t

`t=7.6/9.8=0.776`

This time can be substituted into the height formula:

`h=-1/2 g t^2+v_y t+h_0` sub in values

`h=-1/2(9.8)(0.776)^2+(7.6)(0.776)+0` simplify

`=2.94`

**The height of the centre of the hoop is 2.94 m above the water.**