Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)

A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 68 feet per second. How high will the ball go?

with the information given, you should be able to form the equation

h = -16t^2 + 68t + 6

dh/dt = -32t + 68
= 0 for a max of h
32t = 68
t = 2.125

evaluate h when t = 2.125

To find the maximum height reached by the ball, we can use the equation:

vf^2 = vi^2 + 2aΔy

Where:
vf = final velocity (0 ft/sec, when the ball reaches its maximum height)
vi = initial velocity (68 ft/sec)
a = acceleration (-32 ft/sec^2, due to gravity)
Δy = change in height

Rearranging the equation, we get:

Δy = (vf^2 - vi^2) / (2a)

Substituting the given values, we have:

Δy = (0 - 68^2) / (2 * -32)

Δy = (-4624) / (-64)

Δy = 72 feet

Therefore, the ball will reach a maximum height of 72 feet.

To find the maximum height reached by the ball, we can use the equation of motion for vertical motion:

h(t) = h0 + v0*t + (1/2)*a*t^2

where:
h(t) = height at time t
h0 = initial height
v0 = initial velocity
a = acceleration due to gravity
t = time

In this case, we are given:
h0 = 6 feet
v0 = 68 feet per second
a = -32 feet per second squared (acceleration due to gravity)

Plugging in these values, we have:

h(t) = 6 + 68*t + (1/2)*(-32)*t^2

Since we want to find the maximum height, we need to find the time at which the ball reaches its highest point. At this point, the velocity of the ball will be zero.

To find the time, we can use the fact that the final velocity (vf) equals the initial velocity plus the acceleration multiplied by the time:

vf = v0 + a*t

At the maximum height, the final velocity is zero, so we can solve for t:

0 = 68 - 32*t

Solving this equation, we find:

t = 68/32 = 2.125 seconds

Now we can substitute this value of t into the equation for height to find the maximum height:

h(t) = 6 + 68*2.125 + (1/2)*(-32)*(2.125)^2

Calculating this expression, we get:

h(t) ≈ 6 + 144.5 - 23.081

h(t) ≈ 127.419 feet

Therefore, the ball will go approximately 127.419 feet high.