Question

an object ls launched at 19,6 meters per second from a platform that is 58.8 m tall. h=-4.9t^2 +19.6t +58.8 When does it hit the ground

Answer 1

Answer: the object hits the ground in 6 seconds

Explanation:

t>0

The object hits the ground, i.e. h=0

Hence,

`0=-4.9t^2+19.6t+58.8`

Divide both parts of the equation by -4.9:

0=t²-4t-12

So,

t²-4t-12=0

t²-6t+2t-12=0

(t²-6t)+(2t-12t)=0

t(t-6)+2(t-6)=0

(t-6)(t+2)=0

t-6=0

t=6 seconds

t+2=0

t=-2 seconds ∉ (t>0)