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## 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).