Question

URGENT :

In a shot put event, an athlete throws the shot put from an initial height of 6 feet and with an initial vertical velocity of 29 feet per second. How long until it reaches the ground?

equation is h=-16t^2+29t+6

Answer 1

2

Explanation:

h(t) = -16t² + 29t + 6

h(2) = -16 * 2² + 29 * 2 + 6

h(2) = 0

t = 2 second's

Answer 2

Check the picture below.

so if we just set h = 0, we'll get the "t" when that happened

`\stackrel{h}{0}=-16t^2+29t+6\implies 0=-(16t^2-29t-6)\implies 0= 16t^2-29t-6 \\\\\\ 0=(t-2)(16t+3)\implies t= \begin{cases} ~~ ~ 2 ~~ \checkmark\\ -(3)/(16) ~~ \bigotimes \end{cases}`

now, let's notice that we get two valid values for "t", however the negative doesn't apply in this case, because we can't quite have negative seconds for the object in motion.