Two students are on a balcony 24.3 m above the street. One student throws a ball, b1, vertically downward at 15.8 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 0.510 s after they are thrown?
m
To solve this problem, we need to use the equations of motion for objects in free fall.
(a) The difference in time the balls spend in the air can be calculated by first finding the time of flight for each ball.
For ball b1:
We can use the equation:
h = (1/2) * g * t^2
where h is the height (24.3 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.
Rearranging the equation to solve for t gives:
t = sqrt(2h / g)
t = sqrt(2 * 24.3 / 9.8)
t ≈ 2.78 s
For ball b2:
The time of flight for ball b2 will be the same as ball b1, since they have the same initial speed and same height to cover. Therefore, the time of flight for ball b2 is also approximately 2.78 s.
The difference in time the balls spend in the air is:
Δt = t_b1 - t_b2
Δt = 2.78 s - 2.78 s
Δt = 0 s
Therefore, the difference in time the balls spend in the air is 0 s.
(b) The velocity of each ball as it strikes the ground can be found using the equation:
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 of flight.
For ball b1:
Initial velocity (u_b1) = 15.8 m/s (thrown vertically downward)
Acceleration due to gravity (g) = 9.8 m/s^2
Time of flight (t_b1) = 2.78 s (from part a)
Using the equation, we can calculate the final velocity (v_b1) of ball b1:
v_b1 = u_b1 + gt_b1
v_b1 = 15.8 + (9.8 * 2.78)
v_b1 ≈ 42.68 m/s
For ball b2:
Initial velocity (u_b2) = 15.8 m/s (thrown vertically upward)
Acceleration due to gravity (g) = 9.8 m/s^2
Time of flight (t_b2) = 2.78 s (from part a)
Using the equation, we can calculate the final velocity (v_b2) of ball b2:
v_b2 = u_b2 - gt_b2
v_b2 = 15.8 - (9.8 * 2.78)
v_b2 ≈ -27.92 m/s (negative because it is directed downward)
Therefore, the velocities of the balls as they strike the ground are approximately:
- Ball b1: 42.68 m/s (downward)
- Ball b2: -27.92 m/s (downward)
(c) To find the distance between the balls 0.510 s after they are thrown, we need to find the heights reached by each ball at that time.
For ball b1:
Initial velocity (u_b1) = 15.8 m/s (thrown vertically downward)
Time (t) = 0.510 s
Using the equation:
h_b1 = u_b1 * t + (1/2) * g * t^2
h_b1 = 15.8 * 0.510 + (0.5 * 9.8 * 0.510^2)
h_b1 ≈ 4.122 m
For ball b2:
Initial velocity (u_b2) = 15.8 m/s (thrown vertically upward)
Time (t) = 0.510 s
Using the equation:
h_b2 = u_b2 * t - (1/2) * g * t^2
h_b2 = 15.8 * 0.510 - (0.5 * 9.8 * 0.510^2)
h_b2 ≈ 2.641 m
The distance between the balls 0.510 s after they are thrown is the difference in their heights:
Distance = h_b1 - h_b2
Distance ≈ 4.122 m - 2.641 m
Distance ≈ 1.481 m
Therefore, the balls are approximately 1.481 m apart 0.510 s after they are thrown.