A rock is thrown vertically upward with a speed of 14.0 m/s from the roof of a building that is 50.0 m above the ground. Assume free fall.

In how many seconds after being thrown does the rock strike the ground? in s

What is the speed of the rock just before it strikes the ground? in m/s

Vf^2 = Vo^2 + 2gd,

d = 50 + (Vf^2 - Vo^2) / 2g,
d = 50 + (0 - (14)^2) / -19.6,
d = 50 + 10 = 60m above ground.

Vf = Vo + gt,
t(up) = (Vf - Vo) / a,
t(up) = (0 - 14) / -9.8 = 1.43s.

d = Vo*t + 0.5gt^2 = 60m.
0 + 0.5*9.8t^2 = 60,
4.9t^2 = 60,
t^2 = 12.24,
t(down) = 3.50s.

T = t(up) + t(down),
T = 1.43 + 3.50 = 4.93s

Why did the rock throw a party?

Because it wanted to make a splash!

Now, let's solve the problem. Since the rock is thrown vertically upward and then comes back down, we can use the equation of motion:

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

where h is the height, u is the initial velocity, t is the time, and a is the acceleration (which is equal to -9.8 m/s^2 in this case).

So, let's plug in the values given:

50.0 = 14.0t - (1/2)(9.8)t^2

Rearranging the equation and setting it equal to zero, we have:

(1/2)(9.8)t^2 - 14.0t + 50.0 = 0

Now, we can use the quadratic formula to solve for t. Applying the formula:

t = (-b ± √(b^2 - 4ac))/(2a)

Plugging in the values:

t = (-(-14.0) ± √((-14.0)^2 - 4(1/2)(9.8)(50.0)))/(2(1/2)(9.8))

After evaluating the expression, we find two solutions: t = 4.69 s and t = 10.7 s.

Since the rock is going up and then coming back to the ground, we only consider the positive solution.

So, it takes approximately 4.69 seconds for the rock to strike the ground.

Now, for the speed just before it strikes the ground, we can use another equation of motion:

v = u + at

Plugging in the values:

v = 14.0 - 9.8(4.69)

Calculating the expression, we find that the speed of the rock just before it strikes the ground is approximately 60.64 m/s.

But remember, this is assuming there are no air resistance and no external factors affecting the motion of the rock.

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

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

where:
h = height (in this case, 50.0 m)
v0 = initial velocity (14.0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time

Since the rock is thrown upward, we consider the initial velocity as positive and the acceleration due to gravity as negative.

Plugging in the given values, the equation becomes:

50.0 = 14.0*t + (1/2) * (-9.8) * t^2

Rearranging the equation to get it in standard quadratic form:

-4.9*t^2 + 14.0*t - 50.0 = 0

We can solve this quadratic equation using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

where:
a = -4.9
b = 14.0
c = -50.0

Calculating the values:

t = (-14.0 ± √(14.0^2 - 4*(-4.9)*(-50.0))) / (2*(-4.9))

t = (-14.0 ± √(196 + 980)) / (-9.8)

t = (-14.0 ± √(1176)) / (-9.8)

t = (-14.0 ± √(36 * 4 * 7)) / (-9.8)

t = (-14.0 ± 6√7) / (-9.8)

Since time cannot be negative in this context, we take the positive root:

t = (-14.0 + 6√7) / (-9.8)

Calculating this value:

t ≈ 1.46 seconds

So, the rock strikes the ground approximately 1.46 seconds after being thrown.

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

v = v0 + g*t

where:
v0 = initial velocity (14.0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (1.46 s, as calculated above)

Plugging in the values:

v = 14.0 + (-9.8) * 1.46

v ≈ 14.0 - 14.308

v ≈ -0.308 m/s

Since the velocity is negative, it means the rock is moving downward just before it strikes the ground. The speed, in this case, is the magnitude of the velocity, so the speed of the rock just before it strikes the ground is approximately 0.308 m/s.

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

h = h0 + v0t - (1/2)gt^2

Where:
h = final height (0 m since it hits the ground)
h0 = initial height (50.0 m)
v0 = initial velocity (14.0 m/s)
g = acceleration due to gravity (-9.8 m/s^2, since it is in the downward direction)

Plugging in the values, we have:

0 = 50.0 + (14.0)t - (1/2)(-9.8)t^2

Simplifying the equation, we get:

-4.9t^2 + 14.0t + 50.0 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -4.9, b = 14.0, and c = 50.0. Plugging these values into the quadratic formula, we get:

t = (-14.0 ± √(14.0^2 - 4(-4.9)(50.0))) / (2(-4.9))

Simplifying the equation further, we have:

t = (-14.0 ± √(196 - (-980))) / (-9.8)

t = (-14.0 ± √(196 + 980)) / (-9.8)

t = (-14.0 ± √(1176)) / (-9.8)

Now, we consider the positive root of the equation, since time cannot be negative in this context. So, we have:

t = (-14.0 + √(1176)) / (-9.8)

t ≈ 3.8 seconds

Therefore, it takes approximately 3.8 seconds for the rock to strike the ground.

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

v = v0 + gt

Where:
v = final velocity
v0 = initial velocity (14.0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)

Plugging in the values, we have:

v = 14.0 + (-9.8)(3.8)

v ≈ 14.0 - 37.24

v ≈ -23.24 m/s

The negative sign indicates that the velocity is in the opposite direction to the initial upward velocity. However, since we are only interested in the magnitude, we take the absolute value:

|v| ≈ |-23.24| = 23.24 m/s

Therefore, the speed of the rock just before it strikes the ground is approximately 23.24 m/s.