# A ball is thrown into the air with an upward velocity of 28 ft/s. Its height h in feet after t seconds is [Solved]

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

Time to Max. Ht. = 0.875 sec.

Max. Ht. = 19.25 ft.

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Vertex (0.875, 19.25)

h(t) = 16t 28t 7

h(t) = 16t 28t 7

Convert to vertex type, y = a(x h) ok,
the place (h, ok) is the vertex, h is the time
to max. ht. and ok is the max. ht.:

Vertex (0.875, 19.25)

Vertex (0.875, 19.25)

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

Convert to vertex type, y = a(x h) ok,
the place (h, ok) is the vertex, h is the time
to max. ht. and ok is the max. ht.:

h = -16t^2 28t 7
dh/dt = -32t 28
When h is most, dh/dt = 0.
0 = -32t 28
Fixing t, the worth of t obtained is 0.875.
Due to this fact, the time taken for the ball to succeed in its most peak is 0.875.

Subsequent, to seek out the utmost peak, substitute t=0.875 into the operate h.
Therefore, h = -16(0.875)^2 28(0.875) 7 = 19.25 (nearest hundredth).

Time to Max. Ht. = 0.875 sec.

Max. Ht. = 19.25 ft.

`2/12/14

Convert to vertex type, y = a(x h) ok,
the place (h, ok) is the vertex, h is the time
to max. ht. and ok is the max. ht.:

y = (- 16t 28t) 7
y = 16(t 1.75t) 7
y = 16(t 1.75t 0.7656) 7 [- 16(0.7656)]
y = 16(t 0.875) 7 (- 12.25)
y = 16(t 0.875) 7 12.25
y = 16(t 0.875) 19.25

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

Time to Max. Ht. = 0.875 sec.

Max. Ht. = 19.25 ft.

`2/12/14

This post is last updated on hrtanswers.com at Date : 1st of September – 2022