Question

4. Henry is diving off a diving board that is 7m above the water. After 2 seconds, he reaches his maximum

height of 9m above the water. What height will Henry be after 5 seconds?

Answer 1

Ignore the answer above, the correct answer is 4.5m.

Explanation:

If you ignore all the useless information, you'll see that the parabola Henry creates has a vertex of (2,9) (because obviously the parabola opens downwards). Using this, we can deduce the vertex form of the equation for this parabola to be y=-a(x-2)²+9. This is because the vertex is always equal to (h,k), and the vertex form of any parabola is written as y=a(x-h)²+k. Now, you can solve for a to find it is 0.5, by using (0,7) as the y-intercept (aka the diving board, his starting location). Now, you can graph y=-0.5(x-2)²+9 by using mapping notation, (I find to be the easiest), or any other strategy you'd like. Once graphed, it is clear to see that when the parabola reaches 5 along the x-axis, (simpler if you think of the x-axis as time and the y-axis as height), it is plotted at 4.5 on the y-axis. This means that after 5 seconds, Henry is 4.5 meters above the water.

Answer 2

Explanation:

Given

Diver is standing
`7\ m` above water

Diver is at a maximum height of
`h=7+2=9\ m` at
`t=2\ s`

So, at
`t=2\ s` his velocity is zero

After this he will be under free fall

using

`s=ut+(1)/(2)at^2`

here
`u=0`

`s=(1)/(2)* 10* (3)^2`

`s=5* 9=45\ m\quad \text{(not possible)}`

but the distance between the pool and Henry is
`9\ m`

So, he reaches the pool before
`5\ s`

`9=(1)/(2)* 10* (t)^2`

`t=√(1.8)`

`t=1.34\ s`

So, he reaches pool at
`t=2+1.34=3.34\ s` after diving off the board