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

Answer:

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

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:

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

$v(7.609) = -63.568\,\frac{m}{s}$

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.