Along jumper leaves the ground at 45 ∘ above the horizontal and lands 9.1 m away. what is her takeoff speed v0?

So takeoff speed will be 13.35 m/sec    

Explanation:

We have given that horizontal range R = 9.1 m

Angle from horizontal at which jumper leaves the ground

Acceleration due to gravity

We know that horizontal range is given by

u = 13.35 m/sec

So takeoff speed will be 13.35 m/sec

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.