Think about the oscillation of the toy shown in the video. when the toy is at the highest point of its motion, its velocity is and its acceleration is .

i dont really get what u are asking but here is what im guessing u mean.

Explanation:Refer to the diagram shown below.

m = 50 g = 0.05 kg, the mass of the ball

A = 6 cm = 0.06 m, the amplitude

The oscillatory motion relative to the equilibrium position is

x(t) = A cos(ωt)

where

x =  displacement at time t

A = amplitude, 0.06 m

ω = circular frequency

Therefore

x(t) = A cos(ωt)

The velocity function is

v(t) = - ωA sin(ωt)

The maximum velocity occurs when sin(ωt) = 1. Because the maximum velocity is 3.2 m/s, therefore

(ω 1/s)*(0.06 m) = (3.2 m/s)

w = 53.33 1/s

Therefore

x(t) = 0.06 cos(53.33t)

v(t) = 3.2 sin(53.33t)

When the spring is compressed by 4.5 cm (0.045 m) from the equilibrium position, then

0.06 cos(53.33t) = -0.045

cos(53.33t) = -0.75

53.33t = cos⁻¹ (-0.75) = 2.4189

t = 0.0454 s

The velocity when t = 0.0454 s is

v = 3.2 sin(53.33*0.0454) = 2.111 m/s

The kinetic energy is

KE = (1/2)*m*v²

    = 0.5*(0.05 kg)*(2.111 m/s)²

    = 0.1114 J

 0.1114 J

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be