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

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) (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
---------------------
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
----------------------
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
--------------------
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 (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=245 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.934752415750147, 4.06524758424985. Here's your graph:

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

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