Question

Solve –16t² +144= 0 to find the number of seconds, t, it takes for an object dropped from 144 ft above the ground to hit the ground. Enter the correct values in the boxes to find the solution
$- 16 {t}^{2} + 14 = 0$
$- 16 {t}^{2} + 144 - () = 0 - ()$
$- 16 {t}^{2} = ()$
$\frac{ - 16}{?} {t}^{2} = ()$
${t}^{2} = ()$



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Answer 1

Answer:

h(t) = -16t2 + 144

h(1) = -16(12) + 144 = 128 ft

h(2) = -16(22) + 144 = 80 ft

h(2) - h(1) = 80 - 128 = -48 ft

It fell 48 ft between t = 1 and t = 2 seconds.

It reaches the ground when h(t) = 0

0 = -16t2 + 144

t = √(144/16) s = 3s

It reaches the ground 3s after being dropped.

Step-by-step explanation: