Question #b48bd

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Question #b48bd

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Answer 1 · 2017-06-17T20:57:14

The expression for diver's height above the water line is given as a function of time $t$ $t$

$h (t) = a t^{2} + b t + 10$ $h(t)=at^2+bt+10$ ......(1)

At $t = 0$ $t=0$ , the function reduces to
$h (0) = 10$ $h(0)=10$
Implies that diving platform is located at a height of $10 m$ $10m$ above the water line.
She reaches maximum height of $0.75 m$ $0.75m$ above the platform in $0.5 s$ $0.5s$ . Equation (1) becomes

$h (t) = 10 + 0.75 = a {(0.5)}^{2} + b (0.5) + 10$ $h(t)=10+0.75=a(0.5)^2+b(0.5)+10$
$\Rightarrow 0.75 = 0.25 a + 0.5 b$ $=>0.75=0.25a+0.5b$
Multiplying both sides with $4$ $4$ we get
$a + 2 b = 3$ $a+2b=3$ .......(2)
She reaches back at the platform height in $1 s$ $1s$
Equation (1) becomes
$10 = a + b + 10$ $10=a+b+10$
$\Rightarrow a = - b$ $=>a=-b$ ........(3)
Using (3), equation (2) becomes
$- b + 2 b = 3$ $-b+2b=3$
$b = 3$ $b=3$
and $:.a=-3$

Equation (1) becomes

$h(t)=-3t^2+3t+10$

Comparing with general kinematic expression

$h(t)=1/2at^2+ut+10$
we see that initial velocity of the diver $b=3ms^-1$
and acceleration $2a=-6ms^-2$ .
$-ve$ sign shows that acceleration is acting in a direction opposite to direction of initial velocity.

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Question #b48bd

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