If an object is dropped from a height of 116 feet, the function h(t)=-16t^2+116 gives the height after t seconds. When will the object hit the ground?
let h(t)= 0 (because when it reaches the ground h=0)
0=-16t^2+116
16t^2=116
divide both sides by 16
t^2= 7.25
square root both sides to find t
t=2.69sec
Answer:
2.69 seconds
Step-by-step explanation:
the function for the height and time is:
\(h(t)=-16t^2+116\)
when the object collides with the ground, it means that at that moment its height is zero. So we need to solve the equation above for when h(t) equals zero:
\(0=-16t^2+116\)
clearing for t:
\(-116=-16t^2\\\frac{-116}{-16}=t^2\\ 7.25=t^2\\\sqrt{7.25}=t\\ 2.69=t\)
the object will hit the ground at time 2.69 seconds