So, this is a very interesting question from the projectile motion
Now,
Let us start from the basics,
Considering the boy was at rest while throwing the ball(not running or anything)
So, as the ball is thrown at an at an angle α the horizontal component of velocity will be vcosα and the vertical component will be vsinα
(I am assuming Vo as asked in your question as v)
Let Vx= vcosα
then,
Vy=vsinα
Now, the range of projectile will be ( 2 × Vx × Vy )/g
which will be equal to,
R=( v² × sin2α )/g............... 1
And the time taken will be,
T=(2 × Vx)/g.................2
which will be equal to,
T=( 2× v sinα )/g
So, if he wants to catch the ball then the velocity with which he should run is..
=(Displacement)/(Time)
=Equation 1 divided by equation 2
=(sin2α)/(2sinα)
=(2 ×v× sinα ×cosα )/(2 sinα )
=v cosα
(This line after derivation in short means that horizontal component of the velocity should be the same)
(Since boy was running in horizontally only...so his velocity should be equal to horizontal component of the ball)
