ella_azzopardi
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Figure 2 shows the kite boarder about to perform a jump using a ramp.

The end of the ramp is 1.8 m above the water surface. The kite boarder leaves the ramp at a velocity of 12 m s−1 and at an angle of 17° to the horizontal. The kite boarder lets go of the line at the instant he leaves the ramp.
Calculate the speed with which the kite boarder enters the water.
Assume that the kite boarder is a point mass and ignore the effects of air resistance

Why is this 13.4 and not 6.9?
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Chowderzzz
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would u mind showing your working out? otherwise idk how you'd get 6.9
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ella_azzopardi
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(Original post by Chowderzzz)
would u mind showing your working out? otherwise idk how you'd get 6.9
v is the square root of (3.5^2 + (2 x 9.81 x 1.8)) which is 6.9
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Sinnoh
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(Original post by ella_azzopardi)
v is the square root of (3.5^2 + (2 x 9.81 x 1.8)) which is 6.9
Where are you getting 3.52 from?
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Sinnoh
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(Original post by Sinnoh)
Where are you getting 3.52 from?
Okay I see where you're getting it from, that's the vertical component of the velocity.

The mistake you've made is that the question asks for speed, not vertical velocity - you still have to include the horizontal component of the velocity, which is unchanged.
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