A rock is dropped from a height of 100 meters above the ground. One second later, a ball is thrown vertically downwards, from the same height, with an initial speed of 13 m/s. How long after the ball is thrown will the two objects be at the same location above the ground?

To determine the time when the two objects will be at the same location above the ground, we need to find the time it takes for each object to reach the ground.

Let's start by finding the time it takes for the rock to reach the ground. We can use the equation of motion:

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

where h is the initial height (100m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

Substituting the values, we have:

100 = (1/2) * 9.8 * t^2

Simplifying the equation, we get:

t^2 = (100 * 2) / 9.8

t^2 = 20.408

Taking the square root of both sides, we find:

t ≈ 4.516 seconds

So it takes approximately 4.516 seconds for the rock to reach the ground.

Now let's find the time it takes for the ball to reach the ground. Since the initial downward speed of the ball is 13 m/s, we can use the equation of motion again:

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

Setting h to 100m and v_initial to -13 m/s (negative because it's directed downwards), we have:

100 = (1/2) * 9.8 * t^2 - 13 * t

Rearranging the equation, we get:

4.9 * t^2 - 13 * t + 100 = 0

Solving this quadratic equation gives us two possible solutions, but we only need the positive one since time cannot be negative:

t ≈ 3.961 seconds

So it takes approximately 3.961 seconds for the ball to reach the ground.

Now, to find the time after the ball is thrown when the two objects are at the same location above the ground, we subtract the time it takes for the ball to reach the ground from the time it takes for the rock to reach the ground:

4.516 seconds - 3.961 seconds = 0.555 seconds

Therefore, the two objects will be at the same location above the ground approximately 0.555 seconds after the ball is thrown.

To find out when the two objects will be at the same location above the ground, we need to determine the time it takes for each object to reach the ground.

Let's start with the rock. When an object is dropped, its motion is governed by the equation:

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

where:
h is the height of the object above the ground,
g is the acceleration due to gravity, approximately 9.8 m/s^2,
t is the time.

For the rock, we can use this equation to find the time it takes for the rock to reach the ground:

100 = (1/2) * 9.8 * t^2

Now, let's solve for t:

200 = 9.8 * t^2
t^2 = 200 / 9.8
t^2 ≈ 20.41
t ≈ √20.41
t ≈ 4.52 seconds

So, it takes approximately 4.52 seconds for the rock to reach the ground.

Now let's move on to the ball. The motion of the ball is also governed by the equation:

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

where:
h is the height of the ball above the ground,
v is the initial velocity of the ball,
g is the acceleration due to gravity, approximately 9.8 m/s^2,
t is the time.

For the ball, we know the initial velocity is 13 m/s and the height is 100 meters. The equation becomes:

100 = 13t + (1/2) * 9.8 * t^2

Now let's solve for t:

0 = 4.9t^2 + 13t - 100

Since this is a quadratic equation, we'll solve it using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

For this equation, a = 4.9, b = 13, and c = -100. Plugging these values into the quadratic formula:

t = (-(13) ± √((13)^2 - 4(4.9)(-100))) / (2 * 4.9)

Simplifying further:

t ≈ (-13 ± 21.41) / 9.8

Now we have two possible values for t:
1. t ≈ (-13 + 21.41) / 9.8 ≈ 0.83 seconds
2. t ≈ (-13 - 21.41) / 9.8 ≈ -3.05 seconds (ignoring this negative value)

Therefore, it takes approximately 0.83 seconds for the ball to reach the ground.

Now, to find when the two objects are at the same location above the ground, we need to consider the time difference between them. The ball was thrown 1 second after the rock was dropped. So, we need to add the 1-second delay to the time it takes for the ball to reach the ground:

Total time = 0.83 + 1 ≈ 1.83 seconds

Therefore, approximately 1.83 seconds after the ball is thrown, both objects will be at the same location above the ground.