two students are on a balcony 20.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. (Assume the positive direction is upward.)
(a) What is the difference in the time the balls spend in the air?
s
(b) What is the velocity of each ball as it strikes the ground?
ball thrown downward
m/s
ball thrown upward
m/s
(c) How far apart are the balls 0.800 s after they are thrown?
Thrown down:
From Vf^2 = Vo^2 + 2gh, Vf = 24.89m/s.
From Vf = Vo + gt, t = 1.04sec.
Thrown upward:
From Vf = Vo - gt, t = 1.5sec. to Vf = 0
From h = Vo(t) - gt^2/2, h = 11.025m.
From peak height to ground = 31.625m.
From Vf^2 = Vo^2 + 2gh, Vf = 24.896m/s.
Distance traveled by ball thrown upward after .8 sec. derives from d = Vo(t) - gt^2/2 or d = +8.624m.
Distance traveled by ball thrown downward after .8 sec. derives from d = Vo(t) + gt^2/2 or d = 14.89m.
a) Difference in air time is 4.04 - 1.04 = 3sec.
b) Vf thrown downward = 24.89m/s
...Vf thrown upward = 24.89m/s
c) Therefore, after .8 sec., the balls are 23.52m apart.
To solve this problem, we can use the equations of motion for vertical motion considering the acceleration due to gravity. Let's start by calculating the time each ball spends in the air.
(a) To find the time difference, we need to calculate the time taken by each ball to reach the ground. We can use the equation:
Δy = v₀t + (1/2)gt²
Where:
Δy is the vertical displacement (20.6 m),
v₀ is the initial velocity (14.7 m/s),
t is the time taken,
g is the acceleration due to gravity (-9.8 m/s²).
For the ball thrown downward:
Δy = -20.6 m (taking downward as the negative direction)
-20.6 m = 14.7 m/s * t - (1/2)(9.8 m/s²)t²
Simplifying:
10.9 t² - 14.7 t - 20.6 = 0
We solve this quadratic equation to find the value of t.
Using the quadratic formula: t = (-b ± √(b² - 4ac))/(2a)
t = (-(-14.7) ± √((-14.7)² - 4 * 10.9 * -20.6))/(2 * 10.9)
t ≈ 2.17 s or t ≈ -0.82 s
We neglect the negative value as it does not make physical sense in this context.
Therefore, the time taken by the ball thrown downward to reach the ground is approximately 2.17 s.
For the ball thrown upward, we need to find the time it takes to reach maximum height and then double that time to find the total time in the air since the ball will take the same amount of time to come down as it takes to go up.
Δy = 20.6 m
v₀ = 14.7 m/s
g = -9.8 m/s²
Using the equation:
v = v₀ + gt
0 = 14.7 m/s + (-9.8 m/s²)t
Simplifying:
t ≈ 1.50 s
The time taken for the ball thrown upward to reach maximum height is approximately 1.50 s. Now, to find the total time, we double this value:
2 * 1.50 s = 3.00 s
Therefore, the time taken by the ball thrown upward to reach the ground is approximately 3.00 s.
The difference in the time the balls spend in the air is:
3.00 s - 2.17 s = 0.83 s
Therefore, the difference in time is approximately 0.83 seconds.
(b) To find the velocity of each ball as it strikes the ground, we can use the equation for final velocity:
v = v₀ + gt
For the ball thrown downward:
v = 14.7 m/s - 9.8 m/s² * 2.17 s
v ≈ 14.7 m/s - 21.286 m/s
v ≈ -6.59 m/s (negative sign indicates the direction is downward)
For the ball thrown upward:
v = 14.7 m/s - 9.8 m/s² * 3.00 s
v ≈ 14.7 m/s - 29.4 m/s
v ≈ -14.7 m/s (negative sign indicates the direction is downward)
Therefore, the velocity of the ball thrown downward as it strikes the ground is approximately -6.59 m/s and the velocity of the ball thrown upward as it strikes the ground is approximately -14.7 m/s.
(c) To find out how far apart the balls are 0.800 s after they are thrown, let's calculate the vertical displacement of each ball at that time.
For the ball thrown downward:
Δy = v₀t + (1/2)gt²
Δy = 14.7 m/s * 0.800 s + (1/2)(-9.8 m/s²)(0.800 s)²
Δy = 11.76 m + (-3.92 m)
Δy ≈ 7.84 m (taking upward as the positive direction)
For the ball thrown upward:
Δy = v₀t + (1/2)gt²
Δy = 14.7 m/s * 0.800 s + (1/2)(-9.8 m/s²)(0.800 s)²
Δy = 11.76 m + (-3.92 m)
Δy ≈ 7.84 m (taking upward as the positive direction)
Therefore, the balls are approximately 7.84 meters apart 0.800 seconds after they are thrown.
(a) To find the difference in time that the two balls spend in the air, we need to calculate the time each ball takes to reach the ground. We can use the formula:
time = distance / speed
For the ball thrown downward, the distance is from the balcony to the ground, which is 20.6 m. The speed is given as 14.7 m/s. Plugging the values into the formula:
time downward = 20.6 m / 14.7 m/s = 1.40 s
For the ball thrown upward, the distance is the same, but the speed is also positive. Using the same formula:
time upward = 20.6 m / 14.7 m/s = 1.40 s
Therefore, the difference in time the balls spend in the air is:
time difference = time downward - time upward = 1.40 s - 1.40 s = 0 s
So, the difference in time the balls spend in the air is 0 seconds.
(b) To find the velocity of each ball as it strikes the ground, we need to take into account the acceleration due to gravity. The velocity of an object just before hitting the ground can be found using the formula:
final velocity = initial velocity + (acceleration * time)
For the ball thrown downward, the initial velocity is 14.7 m/s, the acceleration due to gravity is 9.8 m/s² (taking it as negative because it's downward), and the time taken is 1.40 s. Plugging these values into the formula:
final velocity downward = 14.7 m/s + (-9.8 m/s² * 1.40 s) = 1.38 m/s
For the ball thrown upward, the initial velocity is also 14.7 m/s, the acceleration due to gravity is positive this time (since we consider it as upward), and the time taken is 1.40 s. Plugging these values into the formula:
final velocity upward = 14.7 m/s + (9.8 m/s² * 1.40 s) = 28.94 m/s
Therefore, the velocity of the ball thrown downward as it strikes the ground is 1.38 m/s, and the velocity of the ball thrown upward as it strikes the ground is 28.94 m/s.
(c) To find how far apart the balls are 0.800 s after they are thrown, we need to calculate the vertical distance traveled by each ball during that time. We can use the formula:
distance = initial velocity * time + (0.5 * acceleration * time²)
For the ball thrown downward, the initial velocity is 14.7 m/s, the acceleration due to gravity is 9.8 m/s² (taking it as negative), and the time taken is 0.800 s. Plugging these values into the formula:
distance downward = 14.7 m/s * 0.800 s + (0.5 * -9.8 m/s² * (0.800 s)²) = 9.39 m
For the ball thrown upward, the initial velocity is 14.7 m/s, the acceleration due to gravity is positive (since it's upward), and the time taken is 0.800 s. Plugging these values into the formula:
distance upward = 14.7 m/s * 0.800 s + (0.5 * 9.8 m/s² * (0.800 s)²) = 11.23 m
Therefore, the distance between the balls 0.800 s after they are thrown is:
distance apart = distance downward - distance upward = 9.39 m - 11.23 m = -1.84 m
The negative sign indicates that the second ball thrown crossed the first ball and is currently above it.
So, the balls are 1.84 m apart, with the second ball thrown being above the first, 0.800 s after they are thrown.