two children ride bicycle towards each other with instant velocity of 20m/s and 10m/s. Each child releases a ball 6m above the ground with forward horizontal velocity of 40m/s relative to the bicycle. if the two balls collide in above the ground. what was the seperation when they were initially released?
To find the initial separation between the two balls when they were released, we need to determine the time it takes for the balls to collide.
First, let's find the time it takes for each ball to reach the ground. We know that the balls are released 6m above the ground, and the only force acting on the ball is gravity. Using the formula h = (1/2) * g * t^2, where h is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time, we can rearrange the formula to solve for t:
t = sqrt(2h/g)
For the ball released by the first child:
t1 = sqrt(2 * 6 / 9.8) ≈ 0.78 seconds
For the ball released by the second child:
t2 = sqrt(2 * 6 / 9.8) ≈ 0.78 seconds
Since both balls reach the ground at the same time, we can focus on the horizontal motion of the balls.
The horizontal velocity of the first ball relative to the ground is the sum of the velocity of the first child (20 m/s) and the horizontal velocity of the ball relative to the bicycle (40 m/s):
V1 = 20 m/s + 40 m/s = 60 m/s
The horizontal velocity of the second ball relative to the ground is the difference between the velocity of the second child (10 m/s) and the horizontal velocity of the ball relative to the bicycle (40 m/s):
V2 = 10 m/s - 40 m/s = -30 m/s
Now, we can calculate the separation between the two balls when they were initially released. Since both children are riding towards each other, the initial separation is the sum of the distances covered by each ball during the time it takes for the first ball to collide with the second ball.
Separation = V1 * t1 + |V2| * t2
= 60 m/s * 0.78 s + 30 m/s * 0.78 s (taking the absolute value of V2 since we are only interested in distance)
= 46.8 m + 23.4 m
= 70.2 m
Therefore, the initial separation between the two balls when they were released was approximately 70.2 meters.