Question 935429: challenge:
William and Stephanie are playing a rather dangerous game of catch on the rooftops of skyscrapers in downtown Chicago. William is preparing to throw a tennis ball off the roof of the John Hancock Center,1130 feet above street level Stephanie is waiting to catch the ball on the rooftop of the neighboring Water Tower Place,830 feet above street level. William throws the ball,which leaves his hand with an upward velocity of 46 feet per second, and travels in a parabolic path. Stephanie catches the ball exactly 6.0 seconds after it is thrown.
complete the following tasks:
1-Create a quadratic function describing the height of the ball above street level with respect to time t.
2- After how many seconds will the ball reach its maximum height above street level?
3- Find the maximum height of the ball above street level .Round to the nearest foot.
4- Stephanie uses a machine to launch the ball back up to William ,who catches it exactly 6.0 seconds after it is launched. What is the upward velocity in feet per second of the ball as it leaves the launcher? thank you so much.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! William is preparing to throw a tennis ball off the roof of the John Hancock Center,1130 feet above street level
Stephanie is waiting to catch the ball on the rooftop of the neighboring Water Tower Place,830 feet above street level.
William throws the ball,which leaves his hand with an upward velocity of 46 feet per second, and travels in a parabolic path.
Stephanie catches the ball exactly 6.0 seconds after it is thrown.
complete the following tasks:
1-Create a quadratic function describing the height of the ball above street level with respect to time t.
y = -16t^2 + 46t + 1130
Looks like this
green line at 830 ft, intersects at 6 sec
:
2- After how many seconds will the ball reach its maximum height above street level?
max height occurs at the axis of symmetry, x = -b/(2a)
t = 
t = 1.4375 sec for max height
3- Find the maximum height of the ball above street level .Round to the nearest foot.
f(t) = height above street level
f(t) = -16(1.4375^2) 47(1.4375) + 1130
f(t) = -33.0625 + 66.125 + 1130
f(t) = 1163 ft above street level is max height
:
4- Stephanie uses a machine to launch the ball back up to William ,who catches it exactly 6.0 seconds after it is launched.
What is the upward velocity in feet per second of the ball as it leaves the launcher
let v = upward velocity
f(t) = 1130, t = 6 sec
-16(36) + 6v + 830 = 1130
-576 +6v + 830 = 1130
6v = 1130 - 830 + 576
6t = 836
v = 146 ft/sec
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