# Question #b48bd

Jun 17, 2017

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

$h \left(t\right) = a {t}^{2} + b t + 10$ ......(1)

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

$h \left(t\right) = 10 + 0.75 = a {\left(0.5\right)}^{2} + b \left(0.5\right) + 10$
$\implies 0.75 = 0.25 a + 0.5 b$
Multiplying both sides with $4$ we get
$a + 2 b = 3$ .......(2)

3. She reaches back at the platform height in $1 s$
Equation (1) becomes
$10 = a + b + 10$
$\implies a = - b$ ........(3)
4. Using (3), equation (2) becomes
$- b + 2 b = 3$
$b = 3$
and $\therefore a = - 3$

Equation (1) becomes

$h \left(t\right) = - 3 {t}^{2} + 3 t + 10$

Comparing with general kinematic expression

$h \left(t\right) = \frac{1}{2} a {t}^{2} + u t + 10$
we see that initial velocity of the diver $b = 3 m {s}^{-} 1$
and acceleration $2 a = - 6 m {s}^{-} 2$.
$- v e$ sign shows that acceleration is acting in a direction opposite to direction of initial velocity.