Two students are on a balcony 24.3m above the street. One student throws a ball, B1, vertically downward at 12.5 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 individual ball as it strikes the ground?
C.) How far apart are the balls 0.470s after they are thrown?
To solve this problem, we need to consider the motion of the balls individually and analyze their respective trajectories.
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 hit the ground.
For ball B1, which is thrown downward, we can use the equation of motion:
d = v * t + (1/2) * g * t^2
where:
- d is the distance traveled (24.3m in this case),
- v is the initial velocity (-12.5m/s since it's downward),
- t is the time, and
- g is the acceleration due to gravity (-9.8m/s^2).
Plugging in the given values, we have:
24.3 = -12.5 * t + (1/2) * (-9.8) * t^2
Rearranging the equation and solving for t, we get:
4.9t^2 - 12.5t + 24.3 = 0
You can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the coefficients, we have:
t = (-(-12.5) ± √((-12.5)^2 - 4 * 4.9 * 24.3)) / (2 * 4.9)
Solving this equation will give you two possible values for time. However, since we are considering only the positive value of time, we discard the negative value.
B.) Now let's determine the velocity of each ball as it strikes the ground.
For ball B1, we can use the same equation of motion mentioned earlier. Since it is thrown downward, the final velocity before it hits the ground will be negative (indicating downward direction). We can use the equation:
v = v0 + g * t
where:
- v is the final velocity (which we need to find),
- v0 is the initial velocity (-12.5m/s),
- g is the acceleration due to gravity (-9.8m/s^2), and
- t is the time of flight.
Using the positive value of t we obtained earlier, we can substitute it into the equation to calculate v1, the velocity of ball B1 as it strikes the ground.
For ball B2, which is thrown vertically upward, we can use a similar equation:
v = v0 + g * t
In this case:
- v is the final velocity (which we need to find),
- v0 is the initial velocity (12.5m/s upwards),
- g is the acceleration due to gravity (-9.8m/s^2), and
- t is the time of flight.
Substituting the same positive value of t obtained before, we can calculate v2, the velocity of ball B2 as it strikes the ground.
C.) Finally, we can calculate how far apart the balls are 0.470s after they are thrown.
Since B1 is thrown downward and B2 is thrown upward, they have opposite velocities after they are thrown. To find the separation between the two balls at a given time, we can use their velocities and the formula:
s = v1 * t + v2 * t
where:
- s is the separation between the balls at time t,
- v1 is the velocity of ball B1,
- v2 is the velocity of ball B2, and
- t is the given time.
Substituting the values, we can calculate the separation between the balls 0.470s after they are thrown.
By following these steps and solving the equations, you should be able to find the answers to all the questions.