Two students are on a balcony 24.4 m above the street. One student throws a ball, b1, vertically downward at 20.3 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?
(b) What is the velocity of each ball as it strikes the ground?
velocity for b1
velocity for b2
(c) How far apart are the balls 0.510 s after they are thrown?
height of b1:
h1 = 24.4 - 20.3t - 4.9t^2
h1=0 when t = 0.973 sec
h2 = 24.4 + 20.3t - 4.9t^2
h2=0 when t = 5.116 sec
v1 = -20.3 - 9.8t
v2 = 20.3 - 9.8t
plug in the values for t to get the final velocities and heights at 0.510 sec
To solve this problem, we can use the kinematic equations of motion. Let's first calculate the time it takes for both balls to reach the ground.
For ball b1 (thrown downward):
We know that the initial velocity (u) is 20.3 m/s, and the initial displacement (s) is 24.4 m. The acceleration (a) due to gravity is approximately 9.8 m/s².
Using the equation: s = ut + (1/2)at²
24.4 = (20.3)t + (1/2)(-9.8)t²
24.4 = (20.3)t - 4.9t²
After rearranging, we get a quadratic equation: 4.9t² - 20.3t + 24.4 = 0.
We can solve this equation using the quadratic formula. The positive root will give us the time it takes for the ball b1 to reach the ground.
For ball b2 (thrown upward):
The situation is the same as ball b1, but the initial displacement is 0 since it just missed the balcony.
Using the equation: s = ut + (1/2)at²
0 = (20.3)t + (1/2)(-9.8)t²
0 = (20.3)t - 4.9t²
Again, we have a quadratic equation: 4.9t² - 20.3t = 0.
Now, let's calculate the difference in time the balls spend in the air.
(a) To find the difference in time, we subtract the time it takes for ball b2 to reach the ground from the time it takes for ball b1 to reach the ground.
(b) To find the velocity of each ball as it strikes the ground, we can use the equation: v = u + at, where v is the final velocity. For ball b1, u = 20.3 m/s, a = 9.8 m/s², and t is the time calculated earlier. For ball b2, u = 20.3 m/s, a = -9.8 m/s² (opposite direction), and t is the time calculated earlier.
(c) To find how far apart the balls are after 0.510 s, we can use the equation: s = (v1 + v2) × t, where s is the distance between the balls, v1 and v2 are the velocities of the respective balls, and t = 0.510 s.
Now, let's calculate the answers step-by-step.