Two students are on a balcony 21.4 m above the street. One student throws a ball ver- tically downward at 15.6 m/s. At the same instant, the other student throws a ball verti- cally upward at the same speed. The second ball just misses the balcony on the way down.
What is the magnitude of the velocity of the first ball as it strikes the ground?
What is the magnitude of the velocity of the second ball as it strikes the ground?
What is the difference in the time the balls spend in the air?
How far apart are the balls 0.616 s after they are thrown?
first ball distance = -4.9t^2 - 15.6t + 21.4
vel ball1 = -9.8t - 15.5
at ground:
0 = -4.9t^2 - 15.6t + 21.4 or 4.9t^2 + 15.6t - 21.4=0
by formula:
t = 1.035 seconds or some negative t
vel at that time = -9.8(1.035) - 15.5 = -25.64 m/s
magnitude is 25.64 m/s
2nd ball distance = -4.9t^2 + 15.6t + 21.4
repeat the steps I used for ball1
for the last part of the question, sub t = .616 into both equations and subtract their distances.
To solve this problem, we can use the kinematic equations to analyze the motion of the balls.
First, let's find the time it takes for each ball to reach the ground.
For the first ball:
We know the initial vertical velocity (downward) is 15.6 m/s. The final vertical velocity (when it hits the ground) is 0 m/s. The acceleration due to gravity is approximately 9.8 m/s² (assuming negligible air resistance). We can use the kinematic equation:
v_f = v_0 + at
0 m/s = 15.6 m/s + (-9.8 m/s²)t
Simplifying the equation:
-15.6 m/s = -9.8 m/s²t
Solving for t:
t = -15.6 m/s / -9.8 m/s² ≈ 1.59 s
The first ball takes approximately 1.59 seconds to reach the ground.
For the second ball:
The initial vertical velocity is also 15.6 m/s, but this time, it is moving upward. The final vertical velocity is 0 m/s, and acceleration due to gravity is 9.8 m/s². Using the same kinematic equation:
0 m/s = 15.6 m/s + (-9.8 m/s²)t
Simplifying:
-15.6 m/s = -9.8 m/s²t
Solving for t:
t = -15.6 m/s / -9.8 m/s² ≈ 1.59 s
The second ball also takes approximately 1.59 seconds to reach the ground.
Thus, both balls take the same amount of time to reach the ground.
Now, let's find the velocity of each ball when it hits the ground.
For the first ball:
Using the kinematic equation:
v_f = v_0 + at
v_f = 15.6 m/s + (-9.8 m/s²)(1.59 s)
Simplifying the equation:
v_f = 15.6 m/s - 15.6 m/s ≈ 0 m/s
The magnitude of velocity for the first ball as it strikes the ground is approximately 0 m/s since it is moving vertically downward.
For the second ball:
Using the same kinematic equation:
v_f = v_0 + at
v_f = 15.6 m/s + (-9.8 m/s²)(1.59 s)
Simplifying the equation:
v_f = 15.6 m/s - 15.6 m/s ≈ 0 m/s
The magnitude of velocity for the second ball as it strikes the ground is also approximately 0 m/s since it is moving vertically upward.
Thus, both balls have a magnitude of 0 m/s as they strike the ground.
Next, let's find the difference in time the balls spend in the air.
We already determined that both balls take approximately 1.59 seconds to reach the ground. Therefore, the difference in the time the balls spend in the air is 0 seconds since they have the same time of flight.
Finally, let's calculate how far apart the balls are 0.616 seconds after they are thrown.
Since both balls have the same initial vertical velocity of 15.6 m/s but one is moving upward while the other is moving downward, their positions relative to the balcony will be different.
To find the distance between them after 0.616 seconds, we need to calculate the distance each ball has traveled vertically in that time.
For the first ball (moving downward):
Using the kinematic equation:
s = v_0t + (1/2)at^2
s = 15.6 m/s * 0.616 s + (1/2)(-9.8 m/s²)(0.616 s)^2
Simplifying the equation:
s ≈ -4.58 m
The first ball has traveled approximately 4.58 meters downward.
For the second ball (moving upward):
Using the same kinematic equation:
s = v_0t + (1/2)at^2
s = 15.6 m/s * 0.616 s + (1/2)(-9.8 m/s²)(0.616 s)^2
Simplifying the equation:
s ≈ 4.58 m
The second ball has traveled approximately 4.58 meters upward.
To find the horizontal distance between the balls, we need to consider that the balls were thrown simultaneously. Therefore, their horizontal positions are the same. So the distance between them is zero.
In summary:
- The magnitude of the velocity of both balls as they strike the ground is approximately 0 m/s.
- The difference in the time the balls spend in the air is 0 seconds.
- The balls are 0 meters apart after 0.616 seconds since their horizontal positions coincide.