quadratics question, need help please?

Question

A Football player attempts a field goal by kicking the football. The football follows the path modelled by the equation h=3.9t^2+10t+3. where h is the height of the ball and t is the time since the ball was kicked in seconds. the ball must clear the uprights for the field goal to count. the uprights are approx 5 metres high. how long does the ball stay above 5m in height? i'm stumped on this one, any help would be awesome

Answer 1

h = 5m
3.9 t^2 + 10 t + 3 = 5
3.9 t^2 + 10t - 2 = 0
3.9 ( t^2 + 10/3.9 t - 2/3.9) = 0
(t^2 + 10/3.9 t + (5/3.9)^2 - (5/3.9)^2 - 2/3.9) = 0
(t - 5/3.9)^2 - 32.8/15.21 = 0
(t - 5/3.9 - sqrt32.8 / 3.9)(t - 5/3.9 + sqrt32.8 / 3.9) = 0

t = 5/3.9 + sqrt32.8 / 3.9 = 2.75 sec
t = 5/3.9 - sqrt32.8 / 3.9 = -0.187 sec impossible

so time is 2.75 sec

Answer 2

Find out when it is first 5 meters in the air by solving 3.9t^2+10t+3 = 5 or 3.9t^2+10t-2 = 0 by completing the square or the quadratic formula (my choice).

Then find out when it is at it's highest point. That's the vertex (h,k). Solve this by rewriting the equation in vertex form by partially completing the square. Vertex form is y = a (x-h)^2 + k.

Once you know how long it takes before it reaches the highest point and how long it takes to reach the 5 meter mark, you can double the difference in time between these two points. Because if it were to take say 3 seconds to go from 5 meters to it's highest point, it would take 3 seconds for it to come back down to the 5 meter mark too (parabolas are symmetric).

Therefore, the amount of time when it is 5 meters or higher is twice the difference in time between when it reaches the 5 meter mark and when it reaches its maximum height. Good luck!

P.S. I hope noone gives me a thumbs-up for this answer. It's a very silly way of solving the problem. As the first respondent said, just find the difference between the two solutions or roots of 3.9t^2+10t+3 = 5 and you have your answer. I'd go with the quadratic formula as a method of finding these roots.

Answer 3

You want to figure out at what time the ball initially passes 5 meters and the time that it once again falls below 5 meters. To start, you want to solve for 5=3.9t^2+10t+3 or 3.9t^2+10t-2=0. Now just solve this using the quadratic formula: subtract the small root from the large root and that will be your answer.