Two students are on a balcony 21.4 m above the street. One student throws a ball, b1, vertically downward at 13.2 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in time the balls spend in the air?
s
(b) What is the velocity of each ball as it strikes the ground?
velocity for b1 m/s
velocity for b2 m/s
(c) How far apart are the balls 0.570 s after they are thrown?
m
To solve these problems, we can use the equations of motion for objects in free fall. Let's break down each question into steps:
(a) What is the difference in time the balls spend in the air?
Step 1: Calculate the time it takes for the first ball (b1) to hit the ground.
We know that the height (h) from which it is thrown is 21.4 m and the initial velocity (u) is 13.2 m/s in the downward direction. The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Using the formula: h = ut + (1/2)gt^2
Rearranging the equation to solve for time (t), we have:
t = (-u ± √(u^2 - 2gh)) / g
Since the ball is thrown downward, we take the negative sign before the square root.
Substituting the given values:
t = (-13.2 ± √(13.2^2 - 2 * 9.8 * 21.4)) / 9.8
Step 2: Repeat the calculation for the second ball (b2) with the same initial velocity (13.2 m/s) but going upward instead.
Since we know that the second ball will miss the balcony, its time in the air will be longer than the first ball.
After finding both times, we can subtract the time it takes for the first ball to hit the ground from the time it takes for the second ball.
This difference will give us the answer to part (a) - the difference in time the balls spend in the air.
(b) What is the velocity of each ball as it strikes the ground?
Once we have the time taken for each ball to hit the ground, we can use the formula: v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
For the first ball (b1), the initial velocity is 13.2 m/s (downward), and we already know the time it takes to hit the ground.
For the second ball (b2), the initial velocity is also 13.2 m/s (upward), and we have the time it takes for the ball to reach its maximum height. We can assume that the time it takes for the second ball to come back down is twice the time to the maximum height.
By substituting the values into the formula, we can find the velocity of each ball as it strikes the ground.
(c) How far apart are the balls 0.570 s after they are thrown?
To determine the distance between the balls after 0.570 s, we need to calculate the distance each ball has traveled in that time.
For the first ball (b1), we use the formula: s = ut + (1/2)gt^2 to find the distance traveled in 0.570 s.
For the second ball (b2), we need to calculate the time it takes to reach the maximum height and then double that time to find the total time in the air. We can then use the same formula as above to find the distance traveled in 0.570 s.
Finally, we calculate the difference between the distances traveled by the two balls to determine their separation.