A projectile is fired upward from a tower 100 meters high with an initial upward velocity of 98 meters/s...

... Its height above the ground (h) in meters at any time (t) in seconds is given by the equation:

h = -4.9t^2 + 98t + 100

How long (rounded to the nearest hundredth of a second) will it take the projectile to hit the ground?

Please explain your answer - I have to understand this! I think I need to use the quadratic formula... will that work? Thank you so much in advance!!!

Yes, set h = 0 and use the quadratic formula:
x = (-b +/- √(b² - 4ac) )/2a
= (-98 +/- √(98² + 4(4.9)(100)) )/2(-4.9)
= 10/7 (7 +/- √59)
since the value using subtraction is negative, we can ignore it (since t >= 0)

x = 10/7 (7 + √59), which is about 20.97

For a graph, follow the link.

You're looking for the value of t when the projectile reaches the ground (h = 0). First rewrite the equation:

h = 0 = -4.9t² + 98t + 100

4.9t² - 98t - 100

t = [-b ± √(b² - 4ac)] / 2a

Let a = 4.9, b = -98 c = -100. Then:

t = {-(-98) ± √[(-98)² - 4(4.9)(-100)]} / 2(4.9)

t = [98 ± √(9604 + 1960)] / 9.8

t = (98 ± √11564) / 9.8

t = (98 ± 107.5) / 9.8

t = (98 - 107.5) / 9.8 = -0.97

t = (98 + 107.5) / 9.8 = 20.91

The first solution is impossible since it is a negative value. So the solution is that the projectile will hit the ground 20.91 seconds after firing.

Just a side note: There are only two significant digits in this problem, so the solution should actually be 21 seconds.