Maximum Height Word Problem

Question

A pole vaulter leaves the ground, and his height (h) in feet after t seconds is given by

h(t) = -16t² + 19t

a) What was the height of the pole vaulter 1 second after leaving the ground?

b) How many seconds does it take for the pole vaulter to reach his maximum height (to 3 decimal places, if needed)?

c) Use the answer in part (b) to determine the maximum height (to 3 decimal places, if needed)?

Answer 1

h(t) = -16t² + 19t

a). Since this equation promises us that it can be used to calculate the pole vaulter's height as a funtion of time, let's try it by using t=1 sec:

h(t) = -16t² + 19t

h(1) = -16(1)² + 19(1)

h(1) = -16 + 19

h(1) = 3 feet

Sounds reasonable for 1 second, but I don't pole vault.

b). We can find the time to reach maximum height by either of two methods:

1. Differentiate the equation and set it equal to zero. The first derivative tells us the slope of the line formed by this equation (a parabola). Whe the ploe vaulater reaches maximum geight, the slope will be zero for the instant the vaulter's motion turns from positive (goinf up) to negative (going down).
2. Graph the parabola and find the point of maximum height.

Differentiate:

h(t) = -16t² + 19t

h'(t) = -32t + 19

0 = -32t + 19

32t = 19

t = 19/32 sec

Graph:

Plot : There isn't room here for the actual plot. Go to DESMOS.com to produce a graph.

The vertex of the parabola is (19/32, 5.641)

In both cases, the peak height occurs at (19/32) seconds.

c) Use the answer in part (b) to determine the maximum height (to 3 decimal places, if needed)?

Use (19/32) in the original equation:

h(t) = -16t² + 19t

h(19/32) = -16(19/32)² + 19(19/32)

h(19/32) = (361/64)

h(19/32) = 5.641 feet

The pole vaulter reached a maximum height of 5.641 feet after 19/32 seconds.

I suggest additional vaulting practice (if this is correct).