# Converting horizontal speed to vertical speed

• Jul 27th 2006, 06:06 AM
pashah
Converting horizontal speed to vertical speed
Hello I am having some difficulty solving this problem. Any help would be greatly appreciated.

For 1989 and 1990 Dave Johnson had the highest decathlon score in the
world. When Johnson reached a speed of 32ft/sec on the pole vault runway,
his height above the ground t seconds after leaving the ground was given by
h=-16t+32t. (The elasticity of the pole converts the horizontal speed into
vertical speed.) Find the value of t for which his height was 12 ft.

Use the information from the above problem to determine how long Johnson
was in the air. For how long was he more than 14 ft in the air?
• Jul 27th 2006, 06:23 AM
CaptainBlack
Quote:

Originally Posted by pashah
Hello I am having some difficulty solving this problem. Any help would be greatly appreciated.

For 1989 and 1990 Dave Johnson had the highest decathlon score in the
world. When Johnson reached a speed of 32ft/sec on the pole vault runway,
his height above the ground t seconds after leaving the ground was given by
h=-16t+32t. (The elasticity of the pole converts the horizontal speed into
vertical speed.) Find the value of t for which his height was 12 ft.

Use the information from the above problem to determine how long Johnson
was in the air. For how long was he more than 14 ft in the air?

I expect it out to say:

$\displaystyle h=-16t^2+32t$

To find the time (or times) when his height was $\displaystyle 12\text{ ft}$ you
need to solve:

$\displaystyle 12=-16t^2+32t$

or equivalently:

$\displaystyle -16t^2+32t-12=0$.

To find how long he was above $\displaystyle 14\text{ ft}$, first solve:

$\displaystyle -16t^2+32t-14=0$

This has two solutions the first is when he first exceeded $\displaystyle 14\text{ ft}$, and the
second when he subsequently fell below $\displaystyle 14\text{ ft}$. Their difference is how
long he was above $\displaystyle 14\text{ ft}$.

RonL