Is there a mistake in this physics specimen paper question?

I’ve attached it below,(15b)

but essentially how on Earth can 2.22 not be the horizontal velocity component, as the mark scheme says it’s the vertical. And then, they use v-u=at to calculate time, and claim this is the vertical, but how can we know that initial and final velocity are the same but opposite?

?

(edited 8 months ago)

Reply 1

mqb2766

Its parabolic motion, so the vertical velocity is the same magnitude but opposite sign when it returns to the height at which its launched (ground). So the suvat
v^2 = u^2 + 2as
in the vertical direction, s=0 when it returns to the ground so v=+/-u depending on whether t=0 or t=T.

Reply 2

mqb2766

Original post by Nat4695

thank you so much, could you explain why 2.22 is the vertical and not the horizontal component?
i resolved it with 6.5 being the hypotenuse, 20 being the angle between adjacent and hypotenuse as it is to the vertical. this made the opposite side the horizontal and the adjacent the vertical. this gave me 2.22 being the horizontal component of velocity and 6.11 being the vertical

tbh, I misred that. If its 20 to the vertical (usually its to the horizontal) then adjacent is indeed the vertical and youd have 6.5cos(20) for the vertical.

However the mark scheme does the usual 20 to the horizontal and can only presume its a typo in the question.