SOLUTION: URGENT QUESTION NEEDED WITH @ DAYS TIME (Tank you soo much)the question is_ a ball is thrown vertically upwards and its heights at anytime T is given by h=25T-5T^2+1 (h=is in m

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Question 205338: URGENT QUESTION NEEDED WITH @ DAYS TIME (Tank you soo much)the question is_
a ball is thrown vertically upwards and its heights at anytime T is given by
h=25T-5T^2+1
(h=is in metres, T is in seconds)
Find; a) height after 2sec
b)at what time will it reach its maximum height
c)what is the maximum height
d)how long is the ball above 20m
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a ball is thrown vertically upwards and its heights at anytime T is given by
h=25T-5T^2+1
(h=is in metres, T is in seconds)
Find; a) height after 2sec
h(t) = 25t - 5t^2 + 1
h(1) = 25 - 5 + 1
h(1) = 21 meters
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b)at what time will it reach its maximum height
h(t) = 25t - 5t^2 + 1
h(t) = -5t^2 + 25t + 1
Max height is at the vertex, where t = -b/2a
-25/-10 = 2.5 seconds
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c)what is the maximum height
max ht = h(2.5)
h(2.5) = 25*2.5 - 5*2.5^2 + 1
h max = 62.5 - 31.25 + 1
h max = 32.25 meters
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d)how long is the ball above 20m
Find the 2 times where the ball is at 20 m, going up, then going down.
h(t) = 25t - 5t^2 + 1
20 = 25t - 5t^2 + 1
-5t^2 + 25t + 19 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -5x%5E2%2B25x%2B-19+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2825%29%5E2-4%2A-5%2A-19=245.

Discriminant d=245 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-25%2B-sqrt%28+245+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2825%29%2Bsqrt%28+245+%29%29%2F2%5C-5+=+0.934752415750147
x%5B2%5D+=+%28-%2825%29-sqrt%28+245+%29%29%2F2%5C-5+=+4.06524758424985

Quadratic expression -5x%5E2%2B25x%2B-19 can be factored:
-5x%5E2%2B25x%2B-19+=+%28x-0.934752415750147%29%2A%28x-4.06524758424985%29
Again, the answer is: 0.934752415750147, 4.06524758424985. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-5%2Ax%5E2%2B25%2Ax%2B-19+%29

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The time at or above 20 meters is t2-t1
t2 - t1 = 1.4*sqrt(5) seconds