Question

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A diver is standing on a platform 24 ft above a pool. He jumps from the platform with an initial
upward velocity of 8 ft/s. Using the formula, h(t) = -16t2 + vt + s, where h is his height
above the water, t is the time, v is his starting upward velocity, and s is his starting height.
A) After how long will the diver reach its maximum height?
B) What is the maximum height the diver will reach?

Answer 1

Answer:

he Formula h = -16t^2 + vt + s; where

h = height after t seconds

t = time in seconds

v = upward velocity

s = initial height (t=0)

also note that

-16t^2 represents the downward force of gravity for t seconds

+vt = the upward force for t seconds

:

the height when he hits the water = 0 therefore:

0 = -16t^2 + 8t + 24

usually written

-16t^2 + 8t + 24 = 0

we can simplify this, divide by -8, then we have

2t^2 - t - 3 = 0

this will factor to

(2t - 3)(t + 1) = 0

the positive solution

2t = 3

t = 3/2

t = 1.5 seconds to hit the water

-

Answer 2

Form the actual equation

• y=-16t²+vt+s
• y=-16t²+8t+24

Find vertex as vertex is maximum

x coordinate

• -b/2a
• -8/2(-16)
• 8/32
• 1/4s
• 0.25s

y coordinate

• h(0.25)
• -16(0.25)²+8(0.25)+24
• 25m

Max height is 25m

Time to reach is 0.25s