Question #e8c1c

1) If I'm correct, to solve the first question, we set #"H"# equal to #0# because #"H"# is the height of the ball and we are looking for after how many seconds #t# will the ball hit the ground. Well, we know that the ball will be on the ground when the ball is at a height of #0#.

We set up an equation like this and solve for #t#:

#-16t^2+60t+6=0#

You may have noticed that this isn't easy to solve by factoring but we can find the answer by using the quadratic formula:

# t = (-b \pm sqrt(b^2-4ac)) / (2a) #

We let #a=-16,b=60,c=6# and substitute those values into the quadratic formula and solve:

# t = (-(60) \pm sqrt((60)^2-4(-16)(6))) / (2(-16) #

# t = (-60 \pm sqrt(249)) / (-32) #

# t = (15 \pm sqrt(249)) / (8) #

If we evaluate this into a calculator we will get two answers:

#t~~3.87, t~~-0.09#

We can ignore the second answer because since we are talking about time, time can't be negative if we are initially starting when #t=0# seconds. Therefore, the ball will hit the ground at approximately #3.87# seconds (#4# seconds if you want to be loose about it)

2) For this question we simply substitute #3# for #t# to find the height (#"H"#). Thus,

#"H"=-16(3)^2+60(3)+6#

After some algebra...

#"H"=42 "ft"# (after #3# seconds)

P.S. It's good to always visualize the problem. Below is what the path of the ball looked like given by the function in the problem:

graph{-16x^2+60x+6 [-6.7, 153.3, -16.6, 63.4]}