Question

Two students are on a balcony 18.5 m above the street. One student throws a ball ver- tically downward at 13.8 m/s; at the same instant, the other student throws a ball verti- cally upward at the same speed. The second ball just misses the balcony on the way down. How far apart are the balls 0.361 s after they are thrown? The acceleration of gravity is 9.8 m/s2 .

Answer 1

Answer: 11.24m

Step-by-step explanation:

Given that:

• initial velocity of first ball,
  `u_1= 13.8 m.s^(-1)`

• initial velocity of second
  `u_2= -13.8 m.s^(-1)` ( negative because thrown against the gravity)

• time, t = 0.361 s

• acceleration of gravity, g =
  `9.8 m.s^(-2)`

Using the equations of motion:

To find the final velocity after the given time 0.361 second.

`v=u+at` ................ (1)

where:

• v = final velocity of the body after time t under the influence of acceleration a (here g).

Putting the values of the respective balls

`v_1= 13.8+ 9.8 * 0.361`

`v_1= 17.3378 m.s^(-1)`

Similarly

`v_2= -13.8- 9.8 * 0.361` ∵the body is thrown against the gravity will have decelerating effect.

`v_2= -17.3378 m.s^(-1)` ∵the body is still moving up with this velocity

Now, using the eq.

`v^(2)=u^(2)+ 2a.s`.............................(2)

where, the symbols have usual meaning as above and:

• s= distance covered by the body.

Putting the respective values in eq. (2)

For the body in the first case:

`17.3378^(2)= 13.8^(2) + 2 * 9.8 * s`

`s= (17.3378^(2)-13.8^(2))/(2 * 9.8)`

`s = 5.62m`

For the body in the second case:

`(- 17.3378)^(2) = (- 13.8)^(2)- 2 * 9.8* s`

`s= (300.6 - 190.44)/(-2 * 9.8)`

`s=- 5.62m`

∴The bodies are
`2 * 5.62m= 11.24m` apart from each other from the point from which they are thrown.