(h=-16t^2+64t+80), after how many seconds will the object hit the ground at the bottom of the cliff.

1 answer

You must solve equation:

- 16 t ^ 2 + 64 t + 80 = 0

- 16 t ^ 2 + 64 t + 80 = 0 Divide both sides by - 16

- 16 t ^ 2 / - 16 + 64 t / - 16 + 80 / - 16 = 0

t ^ 2 - 4 t - 5 = 0 Add 5 to both sides

t ^ 2 - 4 t - 5 + 5 = 0 + 5

t ^ 2 - 4 t = 5 Add 4 to both sides

t ^ 2 - 4 t + 4 = 5 + 4

t ^ 2 - 4 t + 4 = 9

________________________________________

Remark :

t ^ 2 - 4 t + 4 = ( t - 2 ) ^ 2

________________________________________

( t - 2 ) ^ 2 = 9 Take the square root of both sides

t - 2 = - 3

and

t - 2 = 3

t - 2 = - 3 Add 2 to both sides

t - 2 + 2 = - 3 + 2

t = - 1 s

t - 2 = 3 Add 2 to both sides

t - 2 + 2 = 3 + 2

t = 5 s

Solutions:

t = - 1 a and t = 5 s

Time cant be negative so : t = 5 s

Proof:

- 16 t ^ 2 + 64 t + 80 =

- 16 * 5 ^ 2 + 64 * 5 + 80 =

- 400 + 320 + 80 = 0