At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. Find (a) the time when the balls collide and (b) the height at which they collide. Take g = 10 m/s^2
To find the time when the balls collide, we need to determine when their heights are equal. We'll start by finding the time it takes for each ball to reach the same height.
Let's consider the ball thrown downward first. Using the equation of motion with constant acceleration:
h = vt + (1/2)gt^2
Where:
h = height (40 m),
v = initial velocity (8.0 m/s),
g = acceleration due to gravity (-10 m/s^2),
t = time.
Substituting the given values, we get:
40 = (8.0)t + (1/2)(-10)t^2
Simplifying the equation:
40 = 8t - 5t^2
Rearranging the equation to quadratic form:
5t^2 - 8t + 40 = 0
Using the quadratic formula to solve for t:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 5, b = -8, and c = 40. Plugging these values into the quadratic formula:
t = (-(-8) ± sqrt((-8)^2 - 4 * 5 * 40)) / (2 * 5)
t = (8 ± sqrt(64 - 800)) / 10
t = (8 ± sqrt(-736)) / 10
Since the term inside the square root is negative, we can conclude that the downward ball will never reach the same height as the upward ball since it will hit the ground before they can intersect.
Therefore, there is no time when the balls collide.
As a result, we cannot determine the height at which they collide.