Question

ACTUALLY GUVW ME THE ANSWERS I NEED!!!!
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:

A) t = 0.25 s

B) h = 25 m

Step-by-step 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:

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

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

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