A ball is thrown straight up from the ground at 10m/s, and simultaneously a ball is thrown straight down a 10m-high ledge at 10m/s. At what height do the two balls pass?
To find the height at which the two balls pass, we need to determine the time it takes for each ball to reach that point.
Let's begin with the ball thrown up from the ground. The initial velocity is 10 m/s, and we know that the acceleration due to gravity is approximately 9.8 m/s². Since the ball is moving against gravity's acceleration, we can treat the acceleration as a negative value.
Using the kinematic equation:
h = v₀t + (1/2)at²
where:
h = height
v₀ = initial velocity
t = time
a = acceleration
For the ball thrown up, the values are:
v₀ = 10 m/s (upward)
a = -9.8 m/s² (downward)
Let's set h = 10 m as the distance from the ledge to the ground and solve for t:
10 = 10t - (1/2)(9.8)t²
Simplifying the equation:
-4.9t² + 10t - 10 = 0
Now, let's solve this quadratic equation to find the time it takes for the ball to reach the height of the ledge:
t = (-b ± √(b² - 4ac)) / 2a
a = -4.9, b = 10, c = -10
t = (-(10) ± √((10)² - 4(-4.9)(-10))) / (2(-4.9))
Simplifying further:
t = (-10 ± √(100 - 196)) / -9.8
t = (-10 ± √(-96)) / -9.8
Since we have a negative value under the square root, the equation has no real solutions. This means that the ball thrown up from the ground does not reach a height of 10 m.
Therefore, the two balls do not pass each other at any height.
To determine at what height the two balls pass, we first need to calculate the time it takes for each ball to reach that height.
Let's start with the ball thrown upwards. We know that the initial velocity (u) is 10 m/s, and the acceleration due to gravity (g) is -9.8 m/s² (negative because it acts in the opposite direction of the initial velocity as the ball is thrown upwards).
Using the equation for vertical displacement, we can find the time it takes for the ball to reach its highest point:
u = 10 m/s
a = -9.8 m/s²
v = 0 m/s (velocity at the highest point since the ball is momentarily at rest)
s = ? (height at which the balls pass)
The equation to find the time (t) is:
v = u + at
0 = 10 - 9.8t
Simplifying the equation, we get:
9.8t = 10
Now, solving for t:
t = 10 / 9.8
t ≈ 1.02 seconds
So, the ball thrown upwards takes approximately 1.02 seconds to reach its highest point.
Now let's calculate the time it takes for the ball thrown downwards. Since the ledge is 10 meters high, we can use the equation for vertical displacement again:
s = 10 m
u = 10 m/s (initial velocity)
a = 9.8 m/s² (positive because gravity accelerates the ball downwards)
v = ? (the final velocity is not relevant for this calculation)
Using the equation:
s = ut + (1/2)at²
10 = 10t + (1/2)(9.8)t²
Simplifying the equation, we get:
10t + 4.9t² = 10
Rearranging and simplifying further:
4.9t² + 10t - 10 = 0
This is a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
In this case, a = 4.9, b = 10, and c = -10.
Solving the equation, we get:
t ≈ 0.638 seconds or t ≈ -3.18 seconds
Since time can't be negative in this context, we discard the negative value.
Therefore, the ball thrown downwards takes approximately 0.638 seconds to reach the ledge.
Now, let's calculate the height at which the balls pass.
For the upwards-moving ball, we can use the equation for vertical displacement:
s = ut + (1/2)at²
Substituting the values we know:
s = 10 * 1.02 + (1/2) * (-9.8) * (1.02)²
Simplifying the equation, we get:
s ≈ 5.1 meters
So, the ball thrown upwards reaches a height of approximately 5.1 meters.
Therefore, the two balls pass at a height of approximately 5.1 meters.