A ball is tossed into the air from a bridge, and its height, y (in feet), above the ground t seconds after it is thrown is given by
y = f (t) = −16t^2 + 50t + 36.

a) How high above the ground is the bridge?

b) What is the average velocity of the ball for the first second?

c) Approximate the velocity of the ball at t = 1 second.

d) Graph f , and determine the maximum height the ball reaches. What is the velocity at the time the ball is at the peak?

e) Use the graph to decide at what time, t, the ball reaches its maximum height.

thanks!!!

Question

A ball is tossed into the air from a bridge, and its height, y (in feet), above the ground t seconds after it is thrown is given by
y = f (t) = −16t^2 + 50t + 36.

a) How high above the ground is the bridge?

b) What is the average velocity of the ball for the first second?

c) Approximate the velocity of the ball at t = 1 second.

d) Graph f , and determine the maximum height the ball reaches. What is the velocity at the time the ball is at the peak?

e) Use the graph to decide at what time, t, the ball reaches its maximum height.

thanks!!!

anonymous · Answer

for part a,

i set t to 0 and got 36... is that correct for part a?

anonymous · Answer

yes

anonymous · Answer

okay yay! and for part b i set t to 1 and got -16+50+36=70

and 70 - 36 = 34 ?

is that right for part b? 34?

anonymous · Answer

yea

anonymous · Answer

okay yay!! :)

but i don't understand the rest of the parts... :/ 

could you please explain parts c, d and e to me?

mertsj · Answer

The velocity at t=1 is the slope of the tangent to the curve at t = 1

anonymous · Answer

wait so how do you determine the velocity? do you put all that into an equation? :/

mertsj · Answer

d is the y coordinate of the vertex and the velocity at the vertex is the slope of the tangent to the curve at the  vertex.

mertsj · Answer

wait so how do you determine the velocity? do you put all that into an equation? 
The velocity is the slope of the tangent to the curve at the given point.

mertsj · Answer

e is the x coordinate of the vertex.

anonymous · Answer

okay so would the slope of the tangent be 1 ?

and would the graph look like this? |dw:1391974748892:dw| not really sure :/

anonymous · Answer

and would e be 70?

mertsj · Answer

Don't you have a graphing calculator?

mertsj · Answer

http://www.wolframalpha.com/input/?i=graph%3A+y+%3D+%E2%88%9216x^2+%2B+50x+%2B+36

anonymous · Answer

yes, but i'm confused on what to enter in :/

mertsj · Answer

http://www.wolframalpha.com/input/?i=vertex+of%3A+y+%3D+%E2%88%9216x^2+%2B+50x+%2B+36

anonymous · Answer

ohh okay i see now. you just put in the original equation! i thought you'd have to sub something in! 

okay :)

so with that, e would be 70?

and the graph is d

and c would be 1 ?

mertsj · Answer

yes.  1 is a good estimate for c

anonymous · Answer

okay awesome!! yay!! thank you!!!! :)