Question

The height of an object launched vertically upward is given by the function h(t)=200+11t-4.9t^2, where h is in meters and t is in seconds. What is the object’s approximate speed when it hits the ground?

Answer 1

The approximate speed of the object when it hits the ground is 63.568 meters per second.

Step-by-step explanation:

The height of the object launched vertically upward is represented by the function
`h(t) = 200+11\cdot t -4.9\cdot t^(2)`. The approximate speed of the object when it hits the ground occurs when
`h(t) = 0`. Then, we solve the resulting second-order polynomial by the Quadratic Formula:

`-4.9\cdot t^(2)+11\cdot t +200 = 0` (1)

The roots associated to this polynomial are, respectively:

`t_(1)\approx 7.609\,s` and
`t_(2) \approx -5.364\,s`

Given that time is a positive variable, the only solution that is physically reasonable is:

`t\approx 7.609\,s`

The function velocity is the first derivative of the function defined at the commencement of the explanation, that is:

`v(t) = 11-9.8\cdot t` (2)

If we know that
`t\approx 7.609\,s`, then the approximate velocity of the object when it hits the ground is:

`v(7.609) = 11-9.8\cdot (7.609)`

`v(7.609) = -63.568\,(m)/(s)`

The speed is the magnitude of this velocity. Therefore, the approximate speed of the object when it hits the ground is 63.568 meters per second.