An object is thrown upward at a speed of 121 feet per second by a machine from a height of 8 feet off the ground. The height \( h \) of the object after \( t \) seconds can be found using the equation:

\[ h = -16t^2 + 121t + 8 \]

1. When will the height be 17 feet?

2. When will the object reach the ground?

17=-16t^2+121t+8 16t^2-121t+9=0 this is a quadratic. it solves to 7.487seconds and 0.075 seconds.
0=-16t^2+121t+8 16t^2-121t-8=0 this solves to 7.628 and -0.066. obviously the time cant be negative so the only real answer is 7.628 seconds

Answered by raffi | 2024-06-10

The object will be at a height of 17 feet approximately 0.075 seconds after being launched.
The object will hit the ground approximately 7.63 seconds after being launched.

Answered by IanMckellen | 2024-06-24

The object will reach a height of 17 feet at approximately 0.08 seconds and 7.49 seconds. It will hit the ground at approximately 7.63 seconds after being thrown.
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Answered by raffi | 2024-10-11