Aball player catches a ball 3.13 s after throwing it vertically upward. a)with what speed did he throw it?
b)what height did it reach?

a) v₀ = 15.3 m/s

b) h = 11.9 m

Explanation:

Ball Kinematics

We apply the free fall formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*y Formula (2)

y= v₀t+ (1/2)*a*t² Formula (3)

y:displacement in meters (m)  

v₀: initial speed in m/s  

f: final speed in m/s  

a: acceleration in m/s²  

Data

t= 3.13 s

a=g= -9.8 m/s² = acceleration due to gravity

Problem development

a)With what speed did he throw it

We apply the formula (1) for calculate initial speed (v₀)

vf= v₀+at

0=  v₀+(-9.8)*(3.13/2)

v₀= 15.3 m/s

b)What height did it reach?

We apply formula (2) or formula (3)

vf²=v₀²+2*a*h formula (2)

0 = v₀² +2*(-9.8)*h

19.6*h =  v₀²

h=v₀²/19.6

h= (15.3)²/(19.6 )

h= 11.9m

y= v₀t+ (1/2)*a*t² Formula (3)

h= (15.3)(3.13/2)+(1/2)(-9.8)(3.13/2)²= 11.9m

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one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.