Solving Physics Problems: Ball Throwing Scenario

Two students throwing balls from a balcony

Two students are on a balcony 19.6 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 ver- problems 53 tically upward at the same speed. The second ball just misses the balcony on the way down.

Calculating the difference and velocity of the balls

Answer:
(a) Time difference in ball hitting ground in 2 cases = 3 seconds.
(b) Velocity of both balls hitting ground is same, which is equal to 24.5 m/s.

Explanation:
We have equation of motion, s = ut + ½ at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

a) Time difference:
Let the upward direction be negative and downward direction be positive. Considering the case where the student throws downward:
We have u = 14.7 m/s, acceleration = 9.8 m/s^2, we need to find time when s = 19.6 m.
19.6 = 14.7t + ½ * 9.8 * t^2
Solving the quadratic equation, we find t = 1 second.
Now, considering the case where the student throws upward:
We have u = -14.7 m/s, and following the same calculation, we find t = 4 seconds.
Time difference = 4 - 1 = 3 seconds.

b) Velocity calculation:
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
For the downward throw, the ball strikes the ground at 24.5 m/s after 1 second.
For the upward throw, the ball also strikes the ground at 24.5 m/s after 4 seconds.

Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s, and the other throws a ball upwards at the same speed. What is the difference in the time the two balls take to hit the ground? What is the velocity of each ball as it strikes the ground?

(a) The time difference between the two balls hitting the ground is 3 seconds.
(b) The velocity of both balls as they hit the ground is 24.5 m/s.