SOLUTION: The data in the table shows the height in metres above the ground of a snowball at various times. Showing all work algebraically, create a function that represents the height of th

Although it isn't necessary, it is instructive to
first plot the 5 points with the times on the x-axis
and the heights on the y-axis:

(1,25.1), (2,38.4), (3,41.9), (4,35.6), (5,19.5)

 

It looks as though a parabola would pass through all those points,
so sketch one which passes through all 5 of those points
like this:

 

Now we know that the equation of the height h(t) in meters above the
ground of a projectile is given by this formula:

h(t) = h0 + v0t - 4.9t²

where h0 = the original (initial) height above the
ground that the projectile was thrown, and
v0 = the original (initial) speed in m/s at which
the projectile was thrown upward.

So we substitute the first two points. (1,25.1) and (2,38.4): 

Substituting t=1 and h(1) = 25.1, we have the equation

h(1) = h0 + v0(1) - 4.9(1)²
25.1 = h0 + v0 - 4.9
  30 = h0 + v0

Substituting t=2 and h(2) = 38.4, we have the equation

h(2) = h0 + v0(2) - 4.9(2)²
38.4 = h0 + 2v0 - 4.9(4)
38.4 = h0 + 2v0 - 19.6
  58 = h0 + 2v0

So we have this system of two equations in two
unknowns:

  30 = h0 +  v0
  58 = h0 + 2v0
 
Can you solve that system for h0 and v0?
If not post again asking how.

Answer h0 = 2 meters, v0 = 28 m/s

So substituting in:

h(t) = h0 + v0t - 4.9t²

we find the function h(t) is

h(t) = 2 + 28t - 4.9t²

So really all you needed were two points.
The other three were not necessary, but they
helped make the problem clearer.

Edwin