Stone thrown vertically up at a velocity of 36ft per sec from a bridge 61ft above level water.

Question

H=-16t"2"+36t+61

 

A)Draw a picture

B)Find the time when the stone is again 61 feet above the water. 

C) find the time when the stone strikes the water. (Show answer reduced to lowest term, round nearest thousanth)

Answer 1

Hi Nichole.

 

As for the picture, if you just draw what'd happen in real life, you'd have a straight line going up and down. If you draw the stone's position as a function of time, you'd have an upside-down parabola with the tail end of it stretching further down than the left side of it.

 

To find the time when the stone is 61 feet above the water, set the height H to 61 and solve for t.

 

61 = -16*t^2 + 36t + 61

 

0 = -16t^2 + 36t

 

0 = -4t*(4t - 9)

 

-4t = 0, so t = 0

 

4t - 9 = 0, so t = 9/4 seconds

 

The stone will be 61 feet above the water in 9/4 seconds.

 

To find when the stone hits the water, set H equal to 0 and solve for t

 

0 = -16t^2 + 36t + 61

 

Since these numbers are so ugly (Seriously? 61? Why would the homework say 61...), it'd probably be best to use the quadratic formula.

 

t = (-36 +/- √(1296 +3904))/ -32

 

t = (-36 +/- √(5200))/-32

 

t = (-36 +/- 20√(13))/-32

 

t = (9 +/- 5√(13))/8

 

t = (9 + 5√(13))/8, or

 

t = (9 - 5√(13))/8

 

Since the latter result is negative, and time can't be negative, our answer must be:

 

(9 + 5√(13))/8, or 3.378 seconds

 

Hope this helps.