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A LEVEL PHYSICS - Projectile motion

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.

The mark scheme doesn't make sense - cause I've calculated the correct vertical component (6.9ms-2) but I can't understand how to get the answer 13.4ms-2.

This is the mark scheme:

uh = 12 cos 17° or 11.5 (m s-1) seen
uv = 12 sin 17° or 3.5 (m s-1) seen
Use of v2 = u2 + 2as with either 3.5 or 1.8 or vv = 6.9 (m s-1)
13.4 (m s-1)

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

Reply 2

Henry.m17

You have to times 3.5 by 2 in order to take in account the added height to the ramp, then do Pythagoras with 11.5^2 7^2 rooted to calculate the velocity as 13.5