Question 292003: A rock is thrown upward from the top of a 250-foot cliff with a velocity of 14 feet per second. The height, h, in feet of the object after t seconds is h = –16t2 + 14t + 250. Approximately how long after the object is thrown will it strike the ground below?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A rock is thrown upward from the top of a 250-foot cliff with a velocity of 14 feet per second.
The height, h, in feet of the object after t seconds is h = –16t^2 + 14t + 250.
Approximately how long after the object is thrown will it strike the ground below?
:
Plot this equation:
:
Looks like it will strike the ground in 4.4 sec
:
If you want the exact time, use the quadratic formula to find the positive solution
a=-16; b=14; c=250
:
Response to student comment:
I did use a TI83 to graph the equation, y = -16x^2+14x+250 with a scale:
Window: -2,6,1 and -100,300,20 to get the graph you see
:
But you can graph it on graph paper, by making a table and plotting the x,y points. If you want me to help you do that, my email ankor@att.net
:
What are you confused about?
:
|
|
|
