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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?
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?
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MisterBrainly
4 months ago
Светило науки  2681 ответа  0 помощи
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
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
