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
3 Answers

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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