Projectile motion question?

Question

A football player kicks a football that has a “hang time” of 4.5s and is caught 46m away. If the ball leaves the players foot at a height of 1.5m and is caught overhead at a height of 2.1m, find the ball’s initial velocity.

Answer 1

s=(uy x t)-1/2gt^2 +ho
Where ho= 1.5m and s = 2.1m
2.1=4.5uy- g(4.5^2)/2 +1.5
(2.1-1.5) = 4.5uy- (20.25 x 9.81)/2
0.6 = 4.5uy -99.32
99.32 +0.6 = 4.5uy
uy=99.92/4.5
uy= 22.2 m/s

uy is the vertical component of velocity.
ux is the horizontal component that does not change and is given by ux= 46/4.5= 10.22 m/s

u = squ root of [22.2^2 + 10.22^2]
u = squ root of [598.17]
u =24.4 m/s

Answer 2

Horizontal component of the initial velocity is
(46m)/(4.5s) = 10.2222... m/s.
For the vertical component of motion:
1.5 = 2.1 + vy*4.5 - (1/2)(9.8)(4.5^2) =>
vy*4.5 = -0.6 + 99.225 = 98.625 =>
vy = 21.916666...
So the magnitude of the initial velocity is
sqrt(21.916666...^2 + 10.2222...^2)
= 24.2 m/s.