A rock is thrown vertically upward with a speed of 14.0m/s from the roof of a building that is 70.0m above the ground. Assume free fall.

In how many seconds after being thrown does the rock strike the ground?
What is the speed of the rock just before it strikes the ground?

V = Vo + g*Tr = 0

14 - 9.8Tr = 0
9.8Tr = 14
Tr = 1.43 s. = Rise time.

h = ho + Vo*Tr + 0.5g*Tr^2

hmax = 70 + (14*1.43 -4.9*1.43^2)=80 m.
Above gnd.

hmax = 0.5g*t^2 = 80
4.9t^2 = 80
t^2 = 16.33
Tf = 4.04 s = Fall time.

Tr+Tf = 1.43 + 4.04 = 5.47 s. = Time to
strike gnd.

V=Vo + g*Tf = 0 + 9.8*4.04 = 39.6 m/s.

To find the time it takes for the rock to strike the ground, we can use the equation of motion for vertical motion:

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

Where:
- h is the final height (which is 0 when the rock strikes the ground)
- h0 is the initial height (70.0m)
- v0 is the initial velocity (14.0m/s)
- g is the acceleration due to gravity (-9.8m/s^2)
- t is the time in seconds.

Let's solve the equation for t:

0 = 70.0m + 14.0m/s * t - (1/2) * (-9.8m/s^2) * t^2

This is a quadratic equation in terms of t. We can solve it using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -4.9m/s^2 (1/2 * -9.8), b = 14.0m/s, and c = 70.0m. Plugging these values into the formula:

t = (-14.0m/s ± √(14.0m/s)^2 - 4 * (-4.9m/s^2) * 70.0m) / (2 * -4.9m/s^2)

Simplifying further:

t = (-14.0m/s ± √(196.0m^2/s^2 + 1372.0m^2/s^2)) / (-9.8m/s^2)

t = (-14.0m/s ± √(1568.0m^2/s^2)) / (-9.8m/s^2)

t = (-14.0m/s ± 39.6m/s) / (-9.8m/s^2)

Now we have two solutions for t: one when + is used before 39.6m/s and one when - is used before 39.6m/s. Using the positive solution, we get:

t = (-14.0m/s + 39.6m/s) / (-9.8m/s^2)
t = 25.6m/s / -9.8m/s^2
t ≈ -2.61s (ignoring the negative solution)

Since time cannot be negative, we discard the negative solution. Therefore, the rock will strike the ground approximately 2.61 seconds after being thrown.

To find the speed of the rock just before it strikes the ground, we can use the equation of motion:

v = v0 + g * t

Where:
- v is the final velocity
- v0 is the initial velocity (14.0m/s)
- g is the acceleration due to gravity (-9.8m/s^2)
- t is the time in seconds (2.61s).

Plugging in the values:

v = 14.0m/s + (-9.8m/s^2) * 2.61s

v = 14.0m/s - 25.518m/s

v ≈ -11.518m/s (ignoring the negative sign)

Since the negative sign indicates the direction (downward) and the question asks for the speed, we can ignore the negative sign. Therefore, the speed of the rock just before it strikes the ground is approximately 11.518m/s.