Help needed to find collision force.?

Question

Let says i need to PUNCH a volleyball with the weight of 0.28kg at the angle of 68.8degree with a velocity of 9.5783m/s. Time taken is 0.75s and acceleration is 12.7712m/s^2. The maximum height of the ball must exceed 3.6m and a horizontal distance of 4.6m.
How do i calculate the collision force needed to carry out the above criteria ?

Just to stress on something. It's to PUNCH the ball and not push. So there must be some kind of difference in the calculation from the normal force calculation. Thanks.

Answer 1

F= Fy/sin(A)
F- force in question
Fy - vertical force
A - given angle of 68.8

Fy= mass times change of velocity over time or
Fy=m dV/dt
m - mass of the ball
dV - given as 9.5783 m/s
dt - given as 0.75 s
(A word of caution: we assume that velocity is along the vertical. Is it?)

Acceleration a is given but redundant since
a= dV/dt= 9.5783/0.75= 12.771 m/s^2 (confirmed)

What goes up must comedown and with the same speed

V=at or under gravitational acceleration

V=gt
t=V/g
lets see if it meets the criterion of H=
H= 0.5 gt^2
H=0.5 g(V/g)^2=
H=0.5 V^2 / g=
H=0.5 (9.5783)^2 / 9.81=4.68m
Well it definitely exceeded the required 3.6m

S= Vx t
S - Horizontal distance
Vx - horizontal component of the velocity
t - time to reach the max height.
Oh yes Vx= Vy /tan(A)
Can you figure the rest?