MATH HELP velocity and height?

s = -16t² + v₀t + s₀

Initial velocity of 64 feet per second: v₀ = 64

from the top of 25 ft building: s₀ = 25

To find height equation, simply substitute in the above values:

s = -16t² + 64t + 25

----------

How high is rock after 1 second?

Find s when t = 1

s = -16(1)² + 64(1) + 25

s = 73 feet

----------

After how many seconds will the graph reach max height?

Equation of height is that of a parabola curving down. Maximum height will occur at vertex of this equation. Simply rewrite in vertex form:

s = -16t² + 64t + 25

s = -16 (t² - 4t) + 25

s = -16 (t² - 4t + 4) + 16(4) + 25

s = -16 (t - 2)² + 64 + 25

s = -16 (t - 2)² + 89

Vertex is located at point (2, 89)

Maximum height (or vertex) occurs when t = 2 seconds

----------

What is the maximum height?

According to vertex form, maximum height = 89 feet

____________________ ____________________

NOTE: You can also find time t when rock reaches max height by finding derivative of function (if you've done any calculus yet), setting it equal to 0 and solving for t:

s(t) = -16t² + 64t + 25

s'(t) = -32t + 64 = 0

-32t = -64

t = -64/-32 = 2

max height occurs at time t = 2 seconds

Then to find max height, simply calculate s(2)

s(2) = -16(4) + 64(2) + 25 = -64 + 128 + 25 = 89

max height is 89 feet

initial velocity (v0) is 64 fps

initial height (s0) is 25 ft

s = -16t^2 + vt + s

s = -16 (1)^2 + 64 (1) + 25

s = -16 + 64 + 25

s = 73 feet after 1 second

maximum height is the turning point of the graph

first find axis of symmetry

x = -b/2a

x = -(64) / 2(-16)

x = -64 / -32

x = 2 seconds

Now substitute into original equation to find max height

s = -16 (2)^2 + 64 (2) + 25

s = -16 (4) + 128 + 25

s = -64 + 128 + 25

s = 89 feet

-----------

Calculus approach

Not sure what level you are...

s (t) = -16t^2 + 64t + 25

first derivative

s'(t) = -32t + 64

set equal to zero

0 = -32t + 64

-64 = -32t

2 = t

s(2) = -16 (2)^2 + 64 (2) + 25

s(2) = 89 feet

Second derivative

s''(t) = -32

negative second derivative indicates the graph reaches a maximum value.

this means Vo = 64 and So = 25

so,

s(t) = -16t^2+ 64t + 25

after 1 sec, s(1) = -16 + 64 + 25 = 89 - 16 = 73 ft

for maximum height set ds/dt = 0

ds/dt = -32t + 64 = 0 => t = 2 sec

so s(2) = -16(4) + 64(2) + 25

= 89 ft