Introduction
Two physics students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s, while the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. This scenario presents a classic projectile motion problem in physics which involves determining various kinematic quantities of the thrown balls.
Questions
(a) What is the difference in the two balls' time in the air?
(b) What is the velocity of each ball as it strikes the ground?
(c) How far apart are the balls 0.8 s after they are thrown?
Answers
A) The difference in the two balls' time in the air is approximately 1.1 seconds.
B) The velocity of the first ball as it hits the ground is approximately -33.22 m/s, while for the second ball, it's -15.1 m/s.
C) The distance between the balls 0.8 seconds after being thrown is approximately 23.52 meters.
Explanation
The time for projectile motion is completely determined by the vertical motion. The equation of motion for the ball thrown upward is h = 14.7t - 4.9t^2, while the ball thrown downward has the equation h = -14.7t - 4.9t^2. The time difference in air is approximately 1.1 seconds. To find the velocity of the balls as they strike the ground, use the equation v = u + gt. The negative sign indicates the direction of motion is downwards. To determine how far apart they are 0.8 seconds after being thrown, calculate the position of each ball after 0.8 seconds to find the distance between them.