so, we need to find the solutions for t (the time when the ball is exactly 10 ft in the air). there had to be 2 solutions, as the ball first goes up passing the 10 ft height, and then comes back down again, passing the 10 ft mark a second time. and between these 2 times the ball is higher (but not equal, so, we can only use < or > as inequality signs) than 10 ft.

10 = -t² + 7t

-t² + 7t - 10 = 0

the generation solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -1

b = 7

c = -10

t = (-7 ± sqrt(7² - 4×-1×-10))/(2×-1) =

= (-7 ± sqrt(49 - 40))/-2 = (-7 ± sqrt(9))/-2

t1 = (-7 + 3)/-2 = -4/-2 = 2 seconds

t2 = (-7 - 3)/-2 = -10/-2 = 5 seconds

so, between 2 and 5 seconds airtime the ball is higher than 10 ft.

and remember : HIGHER THAN.

so, we cannot use any equality (like <= or >=).

t must be higher than 2 and lower than 5 :

2 < t < 5

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2 years ago