A stone is tossed from a bridge into the river below. Its path is represented by h = -2t2+ 7t + 9,?

Method 1 : Completing square

a.
h = -2t² + 7t + 9
h = -2[t² + (7/2)t] + 9
h = -2[t² + 2(7/4)t + (7/4)²] + 2*(7/4) + 9
h = -2[t + (7/4)]² + 15.125

As -2[t + (7/4)]² ≤ 0, then h = -2[t + (7/4)]² + 15.125 ≤ 15.125
Hence, the maximum value of h = 15.125
The maximum height = 15.125 m


b.
When h = 15.125 :
-2[t + (7/4)]² = 0
t = 7/4
t = 1.75

Time taken for the stone to reach the maximum height = 1.75 s


c.
When h = 0 :
-2t² + 7t + 9 = 0
2t² - 7t - 9
(2t - 9)(t + 1) = 0
t = 9/2 or t = -1 (rejected)
t = 4.5

Time taken for the stone to reach the water = 4.5 s


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Method 2 : Differentiation

a.
h = -2t² + 7t + 9
h' = -4t + 7
h" = -4

When h = (7/4) :
h = -2(7/4)² + 7(7/4) + 9 = 15.125 m
h' = 0
h' = -4 < 0

Hence, the maximum height = 15.125 m

b.
Refer to part a.
Time taken for the stone to reach the maximum height = (7/4) s = 1.75 s

c.
When h = 0 :
-2t² + 7t + 9 = 0
2t² - 7t - 9
(2t - 9)(t + 1) = 0
t = 9/2 or t = -1 (rejected)
t = 4.5

Time taken for the stone to reach the water = 4.5 s

a. This is found when the slope = 0, so when dh = 0
h' = -4t + 7
0 = -4t + 7
t = 7/4 seconds
t = 1.75 seconds
h = -2(1.75)² + 7(1.75) + 9
h = 15.125 m

b. Shown above

c. Find t when h = 0 m
h = -2t² + 7t + 9
0 = -2t² + 7t + 9
quadratic : (-7 ± √(7² - 4(-2)(9))) / 2(-2)
(-7 ± √(49 +72)) / -4
(-7 ± 11) / -4
t1 = -18 / -4 = 4.5 seconds
t2 = 4 / -4 = -1 second

t2 doesn't work, negative time, so t = 4.5 seconds!

h=-2t^2+7t+9=>
h'=-4t+7
h"=-4<0
h'=0=>
-4t+7=0=>
t=7/4=1.75 s
h(1.75)=15.125 m
(a) The max. h=15.125m
(b) the time needed to reach
the max. h is 1.75 s.
(c) h=0=>
2t^2-7t-9=0=>
t=4.5 s (take +ve t only.
The time needed to reach he water
is 4.5 s.