Question

I got two different answers could someone please help? Mark is in a deep hole looking for treasure. He is standing 10 feet below the surface. He throws an old shoe he found with an initial upward velocity of 36 ft/sec. How long until it lands outside the hole, having gone up and back down? Use the formula -16t^2+36t-10=h , where h is the height of the shoe in feet (relative to the surface) and t is the time in seconds since Mark threw it. Ignore air resistance and round your answer to the nearest tenth.
A.

0.3 seconds
B.

1.9 seconds
C.

2.1 seconds
D.

It will not make

Answer 1

When I did it I got two answers A and B

Answer 2

@phemenway0001

Answer 3

ok so do u know anything about this problem

Answer 4

yes, I know that you plug 0 in for h and then solve, but I got two different answers.

Answer 5

h= -16t^2+36t-10, where h is the height of the shoe in feet (relative to the surface) and t is the time in seconds since Mark threw it. Ignore air resistance and round your answer to the nearest tenth.
h = 0 = -16t^2 + 36t - 10
t^2 - (9/4) t = -5/8
t^2 - (9/4) t + 81/64 = -5/8 + 81/64
(t - 9/8)*2 = 41/64
t = 9/8 +/- sq rt (41/64) = 1.93 sec << answer

Answer 6

Thank you!!

Answer 7

np

Answer 8

Thank you for actually writing it out too!!

Answer 9

np just tag me in any questions u need help with

Answer 10

:)

Answer 11

1.93

Answer 12

\(-16t^2+36t-10=h \)

\(-16t^2+36t-10=0 \)

\(8t^2- 18t + 5=0 \)

\(t = \dfrac{18 \pm \sqrt{(-18)^2 - 4(8)(5)}}{2(8)} \)

\(t = \dfrac{18 \pm \sqrt{324 - 160}}{16} \)

\(t = \dfrac{18 \pm \sqrt{164}}{16} \)

\(t = \dfrac{18 \pm 2\sqrt{41}}{16} \)

\(t = \dfrac{9 \pm \sqrt{41}}{8} \)

\(t = \dfrac{9}{8} + \dfrac{ \sqrt{41}}{8} \) or \(t = \dfrac{9}{8} - \dfrac{ \sqrt{41}}{8} \)

\(t \approx 1.93\) or \(t \approx 0.32\)

The equation does give two solutions, 0.3 sec and 1.9 sec.
Do you understand which one to chose as the correct answer to this problem?