# A ball is thrown into the air with an upward velocity of 28 ft/s. It’s height h in feet after t seconds is?

Given by the function h=-16t^2+28t+7. How long does it take the ball to reach it’s maximum height? What is the ball’s maximum height? Round to the nearest hundredth, if necessary

• a) You can determine the vertical velocity of the ball at any given time by deriving the vertical velocity equation from the vertical displacement equation; you derive according to the power rule:

y = x^n

y’ = nx^(n – 1)

h(t) = (-16t^2 + 28t + 7) m j

v(t) = h'(t) = (-32t + 28) m/s j

When the ball reaches its maximum height, its vertical velocity will be 0 m/s j; therefore,

0 = -32t + 28 => t = 0.875 seconds

b) The ball reaches its maximum height in 0.875 seconds after being thrown; plug and chug and:

h(t) = (-16(0.875)^2 + 28(0.875) + 7) m j = 19.25 m j.

• h = -16t^2 + 28t + 7

dh/dt = -32t + 28

When h is maximum, dh/dt = 0.

0 = -32t + 28

Solving t, the value of t obtained is 0.875.

Therefore, the time taken for the ball to reach its maximum height is 0.875.

Next, to find the maximum height, substitute t=0.875 into the function h.

Hence, h = -16(0.875)^2 + 28(0.875) +7 = 19.25 (nearest hundredth).

• h(t) = – 16t² + 28t + 7

Convert to vertex form, y = a(x – h)² + k,

where (h, k) is the vertex, h is the time

to max. ht. and k is the max. ht.:

y = (- 16t² + 28t) + 7

y = – 16(t² – 1.75t) + 7

y = – 16(t² – 1.75t + 0.7656) + 7 – [- 16(0.7656)]

y = – 16(t – 0.875)² + 7 – (- 12.25)

y = – 16(t – 0.875)² + 7 + 12.25

y = – 16(t – 0.875)² + 19.25

Vertex (0.875, 19.25)

Time to Max. Ht. = 0.875 sec.

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Max. Ht. = 19.25 ft.

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