Answer:
The initial horizontal velocity was 21 m/s
Explanation:
Please, see the figure for a better understanding of the problem.
The equation for the position of an object moving in a parabolic trajectory is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g ·t²)
Where:
r = vector position at time t
x0 = initial horizontal position
v0 = initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity
Notice that at time t = 2.4 s the vector "r" is the one in the figure. We know that the x-component of that vector is 50 m. Then using the equation for the x-component of the vector "r", we can calculate the initial velocity:
x = x0 + v0 · t · cos α
Let´s place the center of the frame of reference at the point of the kick so that x0 = 0.
x = v0 · t · cos α
x/t = v0 · cos α
Notice in the figure that v0 · cos 30° = v0x which is the initial horizontal velocity. Remember trigonometry of right triangles:
cos α = adjacent / hypotenuse = v0x / v0
Then:
50 m/ 2.4 s = v0 · cos 30° = v0x
v0x = 21 m/s