Two students are on a balcony 29.4 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 29.4 m/s . The second ball just misses the balcony on the way down.
a) Sketch the velocity vs. time graph for both balls and paste your sketch in the space below.
b) What is the difference in the time the balls spend in the air?
c) What is the velocity of each ball as it strikes the ground?
d) How far apart are the balls 0.80 sec after they are thrown?
a) To sketch the velocity vs. time graph for both balls, we need to understand the motion of each ball.
The first ball is thrown vertically downwards with an initial velocity of 14.7 m/s. Its velocity will decrease due to the acceleration due to gravity (g = 9.8 m/s^2) until it reaches the ground. The velocity vs. time graph for the first ball will be a straight line sloping downwards.
The second ball is thrown vertically upwards with an initial velocity of 29.4 m/s. Its velocity will also decrease due to the acceleration due to gravity until it reaches its maximum height and starts falling back down. The velocity vs. time graph for the second ball will be a curved line, initially sloping upwards and then sloping downwards.
b) To find the difference in the time the balls spend in the air, we need to calculate the time taken by each ball to reach the ground.
For the first ball, since it is thrown downwards, we can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time taken.
In this case, s = 29.4 m, u = 14.7 m/s, and a = 9.8 m/s^2 (negative since it is in the opposite direction of motion). By substituting these values into the equation, we can solve for t.
For the second ball, since it is thrown upwards and comes down again, we need to find the time taken to reach maximum height and then double it to find the total time spent in the air.
Using the equation v = u + at, we can find the time taken to reach maximum height, where v = 0 m/s. Again, u = 29.4 m/s and a = -9.8 m/s^2 (negative since it opposes the initial velocity).
Then, to find the total time in the air, we double the time taken to reach maximum height.
c) To find the velocity of each ball as it strikes the ground, we need to consider the acceleration due to gravity acting on each ball as they fall.
For the first ball, the velocity as it strikes the ground will be the final velocity, vf = u + at, where u is the initial velocity and a is the acceleration (positive since it is in the same direction as motion).
For the second ball, the velocity as it strikes the ground will also be the final velocity, but in this case, it will be negative since the ball is falling back down. Therefore, vf = -u + at.
By substituting the values of u and a into these equations, we can calculate the velocities.
d) To find how far apart the balls are 0.80 sec after they are thrown, we need to calculate the distance covered by each ball in that time.
For the first ball, we can use the equation s = ut + (1/2)at^2, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time.
Since the first ball is thrown downwards, the initial velocity will be negative. By substituting the values of u, a, and t, we can calculate the distance covered by the first ball in that time.
For the second ball, it is thrown upwards initially. Therefore, we need to calculate the distance covered when the ball is at a certain height above its initial position (in this case, the balcony). Then, when we know this distance, we can calculate the distance from that height to the ground.
Using the equation s = ut + (1/2)at^2, we can find the distance covered by the second ball in the given time.
By calculating these distances, we can find how far apart the balls are.