Math help needed

Question

A ball is thrown from a height of
141
feet with an initial downward velocity of
21/fts
. The ball's height
h
(in feet) after
t
seconds is given by the following.
=h−141−21t-16t^2
How long after the ball is thrown does it hit the ground?

Answer 1

I think the formula could be written as

h(t)=-16t²-21t+141

 

We can always plug the specific values from a story into the general formula h(t)=at²+vt+h₀, where h(t) is the height after t seconds, a is acceleration (gravity), v is velocity, and h₀ is the initial height.

 

We want the height after t seconds to be zero, so we can plug that into the formula also.

 

0=-16t²-21t+141

 

Now we can switch over to the quadratic formula, where a=-16, b=-21, and c=141

 

The quadratic formula say x=(-b±sqrt(b²-4ac))/2a

t=(-(-21)±sqrt(441-4(-16)(141)))/(-32)

t=(21±sqrt(441+9024))/(-32)

t=(21±sqrt(9465))/(-32)

 

It's all computation from there! Any questions?