Answer:
1. 0 seconds.
2. 6.35 seconds.
Step-by-step explanation:
There are two lowest times in the situation, and both occur when the potato is 0 feet from the ground.
0 = -16t^2 + 100t + 10
16t^2 - 100t - 10 = 0
8t^2 - 50t - 5 = 0
To solve for t, use the quadratic formula, where a = 8, b = -50, and c = -5.
[please ignore the A-hat; that is a typo]
[tex]\frac{--50±\sqrt{(-50)^2 - 4 * 8 * (-5)} }{2 * 8}[/tex]
= [tex]\frac{50 ± \sqrt{2500 + 160} }{16}[/tex]
= [tex]\frac{50±\sqrt{2660} }{16}[/tex]
= [tex]\frac{50±\sqrt{2^2 * 5 * 7 * 19} }{16}[/tex]
= [tex]\frac{50 ± 2\sqrt{665} }{16}[/tex]
= [tex]\frac{25±\sqrt{665} }{8}[/tex]
So, one value is...
[25 - sqrt(665)] / 8 = (25 - 25.78759392) / 8 = -0.7875939165 / 8 = -0.0984482396
BUT... the value is negative, and time cannot be negative. So, the LOWEST time that applied to the actual situation is 0 seconds.
The other value is...
[25 + sqrt(665)] / 8 = (25 + 25.78759392) / 8 = 50.78759392 / 8 = 6.34844924
Since the value is positive, it is valid. So, the GREATEST time that applies to the situation is about 6.35 seconds.
Hope this helps!
