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?
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:
*[invoke quadratic "x", 1, -2, -13 ]