A rock is thrown straigh down from a height of 480 feet with an initial velocity of 16 feet per second. How long does it take to hit the ground and with what speed does it hit the ground?

the height h at time t is given by

h(t) = 480 - 16t - 16t^2
when it hits the ground, h=0, so solve
30-t-t^2 = 0
(6+t)(5-t) = 0
t = 5 sec.

the velocity v at time t is

v = -16 - 32t
v(5) = -16 - 160 = -176 ft/s

Well, that rock sure knows how to make a grand entrance! Let's see if we can calculate its timing and speed.

To find out how long it takes for the rock to hit the ground, let's use the kinematic equation for vertical displacement:

S = ut + 0.5gt^2

Where:
S = Vertical displacement (480 feet)
u = Initial velocity (-16 feet per second, as it is thrown downwards)
g = Acceleration due to gravity (-32 feet per second squared)
t = Time taken

Now, let's plug in the values:

480 = -16t + 0.5 * -32 * t^2

When we solve this extremely serious equation, we find that t ≈ 3 seconds.

To determine the final speed of the rock upon hitting the ground, we can use the equation:

v = u + gt

Again, plugging in the values:

v = -16 + (-32) * 3

After some calculations, we find that the rock hits the ground with a speed of approximately -112 feet per second. The negative sign simply indicates that the velocity is downward.

So, there you have it! The rock takes about 3 seconds to hit the ground, and it does so with a humorous -112 feet per second.

To determine the time it takes for the rock to hit the ground, we can use the equation for vertical motion:

h = ut + (1/2)gt^2

where:
h = height (in this case, h = -480 feet since the rock is thrown downwards)
u = initial velocity (u = -16 feet per second since the rock is thrown downwards)
g = acceleration due to gravity (g = 32 feet per second squared)
t = time

Substituting these values into the equation, we have:

-480 = (-16)t + (1/2)(32)t^2

Rearranging the equation, we get:

16t^2 - 16t - 480 = 0

Dividing through by 16, we have:

t^2 - t - 30 = 0

This can be factored as:

(t - 6)(t + 5) = 0

So, t - 6 = 0 or t + 5 = 0. Therefore, the possible solutions are t = 6 seconds or t = -5 seconds. Since time cannot be negative in this context, the time it takes for the rock to hit the ground is t = 6 seconds.

To find the speed at which the rock hits the ground, we can use the equation for velocity:

v = u + gt

where:
v = final velocity (to be determined)
u = initial velocity (u = -16 feet per second)
g = acceleration due to gravity (g = 32 feet per second squared)
t = time (t = 6 seconds)

Substituting these values into the equation, we have:

v = -16 + (32)(6)

v = -16 + 192

v = 176

Therefore, the rock hits the ground with a speed of 176 feet per second.

To find the time it takes for the rock to hit the ground, we can use the kinematic equation:

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

where:
- h is the height of the rock (480 feet)
- vi is the initial velocity (16 feet/second)
- a is the acceleration (due to gravity, which is approximately -32 feet/second^2)
- t is the time we're looking for

Since the rock is thrown straight down, the acceleration is negative.

Plugging in the values into the equation, we get:

480 = 16 * t + (1/2) * (-32) * t^2

Rearranging the equation, we have a quadratic equation:

-16t^2 + 16t + 480 = 0

To solve this equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = -16, b = 16, and c = 480. Plugging in these values, we get:

t = (-16 ± sqrt(16^2 - 4(-16)(480))) / (2(-16))

Simplifying further:

t = (-16 ± sqrt(256 + 30720)) / (-32)
t = (-16 ± sqrt(30976)) / (-32)
t = (-16 ± 176) / (-32)

So, there are two solutions:
t = (16 + 176) / 32 = 192 / 32 = 6 seconds
t = (16 - 176) / 32 = -160 / 32 = -5 seconds

Since time cannot be negative in this context, we discard the negative solution.
Therefore, it takes 6 seconds for the rock to hit the ground.

To find the speed at which the rock hits the ground, we can use another kinematic equation:

v = vi + a * t

Plugging in the values:
v = 16 + (-32) * 6
v = 16 - 192
v = -176 feet per second

The negative sign indicates that the velocity is directed downward, which makes sense since the rock is falling. So, the rock hits the ground with a velocity of 176 feet per second downward.