Question

Question #b48bd

Answer 1

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

`h(t)=at^2+bt+10` ......(1)

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

   `h(t)=10+0.75=a(0.5)^2+b(0.5)+10`
   `=>0.75=0.25a+0.5b`
   Multiplying both sides with `4` we get
   `a+2b=3` .......(2)

3. She reaches back at the platform height in `1s`
   Equation (1) becomes
   `10=a+b+10`
   `=>a=-b` ........(3)
4. Using (3), equation (2) becomes
   `-b+2b=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.