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?
The First Ball:
d = Vo*t + 0.5g*t^2 = 24.4 m.
20.3t + 4.9t^2 = 24.4
4.9t^2 + 20.3t - 24.4 = 0
Use Quadratic Formula:
Tf = 0.973 s.=Fall time or time in air.
The 2nd Ball:
Tr = (V-Vo)/g = (0-20.3)/-9.8 = 2.07 s.=
Rise time.
h = ho + Vo*t + 0.5g*t^2
h=24.4 + 20.3*2.07 - 4.9(2.07)^2=45.4 m
Above gnd.
d = Vo*t + 0.5g*t^2 = 45.4 m.
0 + 4.9t^2 = 45.4
t^2 = 9.27
Tf = 3.04 s.
Tr + Tf = 2.07 + 3.04 = 5.1 s. = Time
in air.
a. 5.1 - 0.973 = 4.14 s.
b. V1^2 = Vo^2 + 2g*d.
V1^2 = (20.3)^2 + 19.6*24.4 = 890.3
V1 = 29.8 m/s.
V2^2 = 0 + 19.6*45.4 = 889.8
V2 = 29.8 m/s.
c. h1 = ho - (Vo*t + o.5g*t^2.)
h1 = 24.4 - (20.3*0.51 + 4.9(0.51)^2
h1 = 24.4 - 11.63 = 12.8 m. Above gnd.
h2 = ho + (20.3*0.51 - 4.9(0.510^2
h2 = 24.4 + 9.08 = 33.5 m. Above gnd.
Da = h2 - h1 = 33.5 - 12.8 = 20.7 m.
Apart.
(a) To find the difference in time the balls spend in the air, we need to determine the time it takes for each ball to reach the ground.
The student throwing ball b1 is throwing it downward, so we can use the equation of motion for free fall:
h = ut + (1/2)gt^2
where h is the height, u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
For ball b1:
h = 24.4 m (height from the balcony to the ground)
u = 20.3 m/s (initial velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
Plugging these values into the equation, we get:
24.4 = 20.3t - (1/2)(9.8)(t^2)
Simplifying the equation:
0 = (1/2)(-9.8)t^2 + 20.3t - 24.4
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
where a, b, and c are coefficients obtained from the equation.
Using the coefficients from our equation:
a = (1/2)(-9.8)
b = 20.3
c = -24.4
Plugging these values into the quadratic formula, we find that t ≈ 2.29 s for b1.
For ball b2, which is thrown upward, the equation of motion is the same:
h = ut + (1/2)gt^2
But this time, the initial velocity is positive since the ball is thrown upward:
u = 20.3 m/s
Using the same equation and plugging in the values:
24.4 = 20.3t - (1/2)(9.8)(t^2)
Simplifying:
0 = (1/2)(-9.8)t^2 + 20.3t - 24.4
Using the quadratic formula, we find that t ≈ 1.06 s for b2.
The difference in time the balls spend in the air is t1 - t2 = 2.29 s - 1.06 s ≈ 1.23 s.
Therefore, the difference in time the balls spend in the air is approximately 1.23 seconds.
(b) To find the velocity of each ball as it strikes the ground, we can use the equation of motion:
v = u + gt
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
For ball b1, the initial velocity is 20.3 m/s (downward), and we can take the time to reach the ground as t ≈ 2.29 s (as calculated earlier).
Plugging these values into the equation, we get:
v1 = 20.3 + (9.8)(2.29)
Simplifying the equation, we find that v1 ≈ -17.2 m/s.
Therefore, the velocity of ball b1 as it strikes the ground is approximately -17.2 m/s (downward).
For ball b2, the initial velocity is 20.3 m/s (upward), and we can take the time to reach the ground as t ≈ 1.06 s (as calculated earlier).
Plugging these values into the equation, we get:
v2 = 20.3 - (9.8)(1.06)
Simplifying the equation, we find that v2 ≈ 10.6 m/s.
Therefore, the velocity of ball b2 as it strikes the ground is approximately 10.6 m/s (upward).
(c) To find how far apart the balls are 0.510 s after they are thrown, we need to determine the distance traveled by each ball in that time.
For ball b1, we can use the equation of motion:
s = ut + (1/2)gt^2
where s is the distance, u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Using the equation and plugging in the values:
s1 = 20.3(0.510) + (1/2)(9.8)(0.510^2)
Simplifying the equation, we find that s1 ≈ 5.12 m.
Therefore, ball b1 travels approximately 5.12 meters in 0.510 seconds.
For ball b2, we can use the same equation of motion:
s = ut + (1/2)gt^2
Using the equation and plugging in the values:
s2 = 20.3(0.510) - (1/2)(9.8)(0.510^2)
Simplifying the equation, we find that s2 ≈ 5.04 m.
Therefore, ball b2 travels approximately 5.04 meters in 0.510 seconds.
Since the two balls are thrown in opposite directions, we can subtract the distances to find the separation between them:
distance = |s1 - s2|
Plugging in the values, we get:
distance = |5.12 - 5.04|
Simplifying the equation, we find that the balls are approximately 0.08 meters (or 80 centimeters) apart after 0.510 seconds.