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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?

User Smisiewicz
by
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1 Answer

4 votes
4 votes

Answer:

A) t = 0.25 s

B) h = 25 m

Explanation:

Given equation:
h(t) = -16t^2 + vt + s

where:

  • h = height above water (in feet)
  • t = time (in seconds)
  • v = initial upwards velocity = 8 ft/s
  • s = initial height = 24 ft

A) To find the value of t when h is at its maximum, differentiate the equation with respect to t:


\implies h'(t)=-32t+v

Substitute given value of v:


\implies h'(t)=-32t+8

Set to zero and solve for t:


\implies -32t+8=0


\implies t=0.25 \textsf{s}

B) Substitute the found value for t into the equation and solve for h:


\implies h(0.25) = -16(0.25)^2 + 8(0.25) + 24


\implies h(0.25) = -1 + 2 + 24


\implies h(0.25) =25 \textsf{ m}

User Techflash
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3.1k points