Question

H=-16t^2+46t+6 after how many seconds will the ball hit the ground?

Answer 1

Ball hit the ground is when the function equals zero
0=-16t^2+46t+6
we can't factor so use quadraticformula

for
ax^2+bx+c=0
x=$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
x=$\frac{-46+/- \sqrt{46^2-4(-16)(6)} }{2(-16)}$
x=$\frac{-46+/- \sqrt{2116+384} }{-32}$
x=$\frac{-46+/- \sqrt{2500} }{-32}$
x=$\frac{-46+/- 50 }{-32}$
x=$\frac{-46+ 50 }{-32}$ or $\frac{-46- 50 }{-32}$
x=$\frac{4}{-32}$ or x=$\frac{-96}{-32}$
x=-1/6 or x=3

we cannot have negative time because time starts at 0
disregard -1/6 as a possible answer


answer is after 3 seconds