SOLUTION: Bob throws a ball vertically upward from the top of the cliff, the height of the ball above the base of the cliff is approximated by the model h = 65 + 10t- 5t ^2 where h is the he

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Question 694657: Bob throws a ball vertically upward from the top of the cliff, the height of the ball above the base of the cliff is approximated by the model h = 65 + 10t- 5t ^2 where h is the height in metres and t is the time in seconds.
a) How high is the cliff?
b) How long does it take the ball to reach a height of 50 metres above the cliff?
c) After how many seconds does the ball hit the ground?
can you please help me out? I really don't understand can you go step by step. Thanks in advance :)
Found 2 solutions by nerdybill, DrBeeee:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Bob throws a ball vertically upward from the top of the cliff, the height of the ball above the base of the cliff is approximated by the model h = 65 + 10t- 5t ^2 where h is the height in metres and t is the time in seconds.
a) How high is the cliff?
h = 65 + 10t- 5t ^2
set t to zero to find height of cliff:
h = 65 + 10(0)- 5(0)^2
h = 65 metres
.
b) How long does it take the ball to reach a height of 50 metres above the cliff?
set h to 50 and solve for t
h = 65 + 10t- 5t ^2
50 = 65 + 10t- 5t^2
0 = 15 + 10t- 5t^2
0 = -5t^2+10t+15
0 = 5t^2-10t-15
0 = t^2-2t-3
0 = (t+1)(t-3)
t = {-1, 3}
throw out the negative solution leaving:
t = 3 seconds
.
c) After how many seconds does the ball hit the ground?
set h to 0 and solve for t:
h = 65 + 10t- 5t ^2
0 = 65 + 10t- 5t^2
0 = -5t^2+10t+65
0 = 5t^2-10t-65
0 = t^2-2t-13
solve using the "quadratic equation" to get
t = {4.74, -2.74}
throw out the negative solution leaving:
t = 4.74 seconds
.
Details of quadratic equation follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-13+=+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=%28-2%29%5E2-4%2A1%2A-13=56.

Discriminant d=56 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+56+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+56+%29%29%2F2%5C1+=+4.74165738677394
x%5B2%5D+=+%28-%28-2%29-sqrt%28+56+%29%29%2F2%5C1+=+-2.74165738677394

Quadratic expression 1x%5E2%2B-2x%2B-13 can be factored:
1x%5E2%2B-2x%2B-13+=+1%28x-4.74165738677394%29%2A%28x--2.74165738677394%29
Again, the answer is: 4.74165738677394, -2.74165738677394. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-13+%29

Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
That's better!
We are given
(1) h = -5t^2 + 10t +65
To answer a) just let t=0 and get
(2) h = 0 + 0 + 65 or
(3) h = 65 m
This is the height of the cliff he is standing on before he throws the ball, t=0. Actually he would be holding the ball in his hand which is above the cliff, but we can't figure this out from the data given, so we assume the cliff itself is 65 m. high.
To answer b) we need another clarification of the problem, because the ball reaches a maximum height of 70m in one second. So question b) must want the time it takes the ball to reach 50m above the base of the cliff NOT above the cliff. Assuming this is what is meant we set h = 50 and solve for t,
(4) -5t^2 + 10t + 65 = 50 or
(5) -5t^2 + 10t + 15 = 0 or dividing through by (-5) gives us
(6) t^2 - 2t - 3 = 0 which factors to
(7) (t - 3)*(t + 1) = 0 giving a pair of roots
(8) t = (3,-1) from which we select the positive time of 3s
Let's check this value using (1)
Is (-5*3^2 + 10*3 +65 = 50)?
Is (-45 + 30 + 65 = 50)?
Is (95 - 45 = 50)?
Is (50 = 50)? Yes
Answer to b) It takes 3 second for the ball to reach a height of 50m above the base of the cliff.
Now for question c), we do the same as in the answer to b) except we set h equal to zero and solve for t,
(9) -5t^2 + 10t + 65 = 0, again dividing through by (-5) simplifies (9) to
(10) t^2 - 2t - 13 = 0 which gives two roots (use the quadratic formula)
(11) t = (1+sqrt%2814%29, 1-sqrt%2814%29)
Select the positive value of time gives
(12) t = 1 + sqrt%2814%29
Use (1) to check this answer.
Is (-5*(1+sqrt%2814%29)^2 + 10*1+sqrt%2814%29) + 65 = 0)?
Is (-5*(1+2*sqrt%2814%29+14) + 10*(1+sqrt%2814%29) +65 = 0)?
Is (-5 -10*sqrt%2814%29 -70 + 10 + 10*sqrt%2814%29 +65 = 0)?
Is (-5 -70 +10 +65 = 0)?
Is(0 = 0)? Yes
Answer to c): The ball takes 1+sqrt%2814%29 (approx 4.74) seconds to reach the base of the cliff (hit the ground).