Projectile Motion of broad jumper

[SigFig]

Homework Statement

A broad jumper begins his jump at an angle of 22 degrees from the horizontal. At the middle of the jump he reaches a height of 70.0cm
a. What is his horizontal velocity
b. What is his horizontal displacement
ANSWER: 9.16m/s

Homework Equations

dh=vht
dv=1/2at^2
t=vv/a

The Attempt at a Solution

Sin 22 = x/70 x= 26.2
26.2=1/2(9.8)t^2
√t
t= 2.3sec

Since it's half way up, you'd have to double the time to find the entire time(?) needed for the entire horizontal displacement, but I have no clue what to do after getting time (if that's even right).

 

[haruspex]

Homework Statement

A broad jumper begins his jump at an angle of 22 degrees from the horizontal. At the middle of the jump he reaches a height of 70.0cm
a. What is his horizontal velocity
b. What is his horizontal displacement
ANSWER: 9.16m/s

Homework Equations

dh=vht
dv=1/2at^2
t=vv/a

The Attempt at a Solution

Sin 22 = x/70 x= 26.2

No, the angle is his initial trajectory. It does not provide a straight line to his highest point.
What vertical component of speed would he need at take-off to reach that height?
What, therefore, was his horizontal speed?
Btw, it's not clear to me whether "horizontal displacement" refers to where he lands or at the highest point. I think it's the highest point.