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 find out when the two objects will be at the same location above the ground, we need to calculate how long 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 for free fall:
s = ut + (1/2)gt^2
Where:
s = distance traveled (in this case, 100 meters)
u = initial velocity (0 m/s, as the rock is dropped)
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values, we get:
100 = 0t + (1/2)(9.8)t^2
Simplifying the equation, we have:
4.9t^2 = 100
Dividing both sides by 4.9, we get:
t^2 = 100 / 4.9
t^2 ≈ 20.41
Taking the square root of both sides, we find:
t ≈ √20.41
t ≈ 4.52 seconds (rounded to two decimal places)
Therefore, it takes approximately 4.52 seconds for the rock to reach the ground.
Next, let's calculate when the ball will reach the ground. Since it is thrown downward, we can use the equation for vertical motion:
s = ut + (1/2)gt^2
Where:
s = distance traveled (in this case, 100 meters)
u = initial velocity (-13 m/s, as it is thrown downward)
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values, we get:
100 = -13t + (1/2)(9.8)t^2
Rearranging the equation, we have:
4.9t^2 - 13t + 100 = 0
Solving this quadratic equation gives us two possible values for t. However, we are only interested in the positive time since the ball is thrown downward.
Using the quadratic formula, we find:
t ≈ 5.23 seconds (rounded to two decimal places)
Therefore, it takes approximately 5.23 seconds for the ball to reach the ground.
Since we are interested in the time after the ball is thrown, we need to find the difference between the time it takes for the ball to reach the ground and the time it takes for the rock to reach the ground:
Time after the ball is thrown = Time for the ball to reach the ground - Time for the rock to reach the ground
Time after the ball is thrown ≈ 5.23 seconds - 4.52 seconds
Time after the ball is thrown ≈ 0.71 seconds (rounded to two decimal places)
Therefore, the two objects will be at the same location above the ground approximately 0.71 seconds after the ball is thrown.