Help solving the following word problem...

[Steph Annie]

Hello!

Could you help me with the following word problem? Thanks...

"A ball thrown off the top of a building has a height h(t) in feet given by the function h(t) = -16t^2 + 80t +96 where t is the number of seconds after it is thrown.

a) How high is the ball after 3 seconds?

This is what I did: h(3) = -16(3)^2 + 80 (3) + 96
3h = -144 +240 + 96
3h=192
h= 192/3

h = 64, The ball is 64ft high after 3 seconds.

Is this work right? I'm not sure how to do the rest...the equation part below especially.

b) When does the ball hit the ground? Write and equation to solve this problem and solve it.

c) What is the maximum height reached by the ball?

 

[Steph Annie]

Ok. So, h(3) = -16(3)² + 80(3)+96
h = -144 + 240 +96
h = 192

When does the ball hit the ground? Write an equation to solve this problem and solve it...

h(t)=0

-16t²+80t +96 =0
t²-5t-6=0
(t-6)(t+1) = 0
t=6 (since t can't be -1)

Is that correct?

And what does the t represent again? Would I say the ball hits the ground after 6 seconds?

Then what do I do for c) maximum height reached by the ball?

wat? Why 3h? h is a function of t, not a constant. you want to write

h(3) = -144 +240 + 96 = 192

i.e. at 3 seconds the ball is 192 feet high.



What is the height of the ground? set h(t) equal to that height and solve. One solution won't make sense and one will.



what is the maximum value reached by h(t) ? It's a down facing parabola so it has some top point. You should be able to solve for that.

Try getting h(t) into standard parabola form A (x - x0)² + B

That will help you see what it's maximum value is.

 

[Steph Annie]

Great. And standard parabola form is y= ax^x+bx +c
Correct?

And again, I am putting the equation I made in....So, this would be the equation t²-5t-6=0
Right?

How do I put it into the parabola form? Sorry to ask but I'm just not visualizing it...

yes it is. nicely done.



yes. You might wonder what the -1 is. Basically what's going on is that you start at a height of 96 ft, with a velocity of 80 ft/s upwards. The t² term is the acceleration caused by gravity downwards. If you track the trajectory back the other way, backwards in time, you'd see the thing was launched from the ground at t = -1s.



convert your equation into standard parabola form. This will let you read off the maximum easily.