Question

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

GIVEN ↷

• h(t) = -16t² + vt + s

where,

h⤿ Height ⤏0

t ⤿ time⤏?

v⤿initial velocity⤏8ft/s

s⤿initial height⤏24ft

TO FIND ↷

• The maximum height
• Time taken to reach the maximum height

SOLUTION ↷

we know that,

• -16t² + vt + s = h(t)
• -16t² + 8t + 24 = 0

To factorise the equation, we'll divide the both side by 8 and multiply by -1

we get,

• 2t² -t -3 = 0
• 2t² +2t -3t -3 = 0
• 2t(t+1) -3(t+1) = 0
• (2t-3)(t +1) = 0

so,

2t - 3 = 0

2t = 3

t = 3/2

t = 1.5 seconds

hence, the diver will take 1.5 seconds to reach the maximum height

Now , to find the maximum height ,

we'll insert the value of time taken in the given equation which is ↷

• h(t) = -16t² + vt + s
• h(1.5) = -16(1.5)² + 8 x 1.5 + 24
• h(1.5) = -16 x 2.25 + 12 +24
• h(1.5) = -36 +12 +24
• h(1.5) = -36 +36
• h(1.5) = 0
• h = 0/ 1.5
• h = 0

Hence , the maximum height the diver will reach in 1.5 seconds = 0