Question

A diver launches herself off a springboard. The height of the diver, in meters, above the pool t seconds after launch can be modelled by the following function:
$h(t)=5t-10t^{2}+10$, t≥0

a) How high is the springboard above the water?
b) Use the model to find the time at which the diver hits the water.
c) Rearrange h(t) into the form A-B(t-C)^2 and give the values of the constants A, B and C.
d) Using your answer to part c or otherwise, find the maximum height of the diver, and the time at which this maximum height is reached.

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Here The height of the diver above the pool t seconds after launch can be modelled by the function :

$\sf{h(t) = 5t - 10 {t}^{2} + 10} \: \: \: ....(1)$

a) The height of the springboard above the water is obtained by putting t = 0 in Equation (1)

Which gives

$\sf{h(0) = 5 \times 0 - 10 \times {(0)}^{2} + 10} \: = 10$

Hence the height of the springboard above the water is 10 metre

b) The time at which the diver hits the water is obtained by solving

$\sf{h(t) =0}$

$\implies \: \sf{ 5t - 10 {t}^{2} + 10} = 0$

$\implies \: \sf{ 10 {t}^{2} - 5t - 10\: } = 0$

$\implies \: \displaystyle \sf{ t = \: \frac{5 \: \pm \: \sqrt{25 + (4 \times 10 \times 10)} }{2 \times 10} }$

$\implies \displaystyle \sf{ t = \frac{5 \: \pm \: \sqrt{425} }{20 } }$

Since time ( t ) cannot be negative

So

$\implies \displaystyle \sf{ t \ne \: \frac{5 \: - \: \sqrt{425} }{20 } }$

So

$\implies \displaystyle \sf{ t = \: \frac{5 + \sqrt{425} }{20 } } = 1.28$

Hence the time at which the diver hits the water is 1.28 seconds

c) The given equation is

$\sf{h(t) = 5t - 10 {t}^{2} + 10} \:$

$\implies \sf{h(t) = 5t - 10 {t}^{2} + 10} \:$

$\implies \sf{h(t) = - 10 {t}^{2} + 5t+ 10} \:$

$\displaystyle\implies \sf{h(t) = - 10( {t}^{2} - \frac{t}{2} ) + 10} \:$

$\displaystyle\implies \sf{h(t) = - 10( {t}^{2} - \frac{t}{2} + \frac{1}{16} ) + 10 + \frac{10}{16} } \:$

$\displaystyle\implies \sf{h(t) = - 10{(t - \frac{1}{4}) }^{2} + \frac{170}{16} } \:$

$\displaystyle\implies \sf{h(t) =\frac{85}{8} \: - 10{(t - \frac{1}{4}) }^{2} }$

Which is of the form

$\sf{h(t) = A-B{(t-C)}^{2} }$

$\displaystyle \sf{ \: }Where \: \: \: A = \frac{85}{8} \: , \: B = - 10 \: , \: C = \frac{1}{4}$

d) Thus the given equation

$\sf{h(t) = 5t - 10 {t}^{2} + 10} \: \: \:$

can be rewritten in the form

$\displaystyle \sf{h(t) =\frac{85}{8} \: - 10{(t - \frac{1}{4}) }^{2} }$

As we know square of a real number can not be negative

$\displaystyle \sf{Hence \: maximum \: value \: of \: \: h(t) \: = h_{max}=\frac{85}{8} }$

This happens when

$\displaystyle \sf{{t - \frac{1}{4} } = 0}$

$\implies \: \displaystyle \sf{{t = \frac{1}{4} } }$

$\displaystyle \sf{} \: Hence \: the \: maximum \: height \: of \: the \: diver \: = \frac{85}{8} \: \: \: metre$

$\displaystyle \sf{} The \: time \: at \: which \: this \: maximum \: height \: is \: reached \: is \: \frac{1}{4} \: \: \: seconds$

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Answer 2

Answer:

a) The springboard is at a height of 10 unit.

b) The time at which the diver hits the water =  $\frac{1 - \sqrt{-15} }{4}$ unit.

c) h(t)  = $\frac{85}{8}$ - 10(t - $\frac{1}{4}$ )²    and   A =  $\frac{85}{8}$ ,  B = 10,  C =  $\frac{1}{4}$

d) The maximum height of the diver = $\frac{1}{4}$ unit

    The time at which this maximum height is reached =   $\frac{85}{8}$ unit.

Step-by-step explanation:

Height, h(t) = 5t - 10t² + 10

(a) When the springboard is above water then the diver launches herself so at that time t = 0.

h(0) = (5×0) - (10×0) + 10 = 10 unit

∴ The springboard is at a height of 10 unit.

(b) When the diver hits the water then there is no distance between water and springboard. So h = 0.

                                         ⇒ 5t - 10t² + 10 = 0

                                         ⇒  5(t - 2t² + 2) = 0

                                         ⇒  t - 2t² + 2 = 0

                                          ⇒ 2t² - t - 2 = 0

                                          ⇒   t = $\frac{1 - \sqrt{-15} }{4}$ (as time can not be negative)

  ∴The time at which the diver hits the water =  $\frac{1 - \sqrt{-15} }{4}$ unit.

(c)  h(t) = 5t - 10t² + 10

           = - 10(t² - $\frac{t}{2}$ ) + 10

           = - 10(t² - $\frac{t}{2}$ + $\frac{1}{16}$) + 10 + $\frac{10}{16}$

           =  - 10(t² - 2*$\frac{t}{2}$*$\frac{1}{4}$ + $\frac{1}{4^2}$) + 10 + $\frac{10}{16}$

           = - 10(t - $\frac{1}{4}$ )² + $\frac{170}{16}$

           = $\frac{85}{8}$ - 10(t - $\frac{1}{4}$ )².............(1)

T given format is,  h(t) = A - B(t-C)²................(2)

Now comparing (1) and (2), A =  $\frac{85}{8}$

                                             B = 10

                                             C =  $\frac{1}{4}$

d) From equation(1) we get h(t)  = $\frac{85}{8}$ - 10(t - $\frac{1}{4}$ )²

  (t - $\frac{1}{4}$ )² = always positive.

  h(t) will be maximum when (t - $\frac{1}{4}$ )² = 0 that means t = $\frac{1}{4}$ unit

   and maximum height = $\frac{85}{8}$ unit.