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

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

$v_1= 13.8+ 9.8 \times 0.361$

$v_1= 17.3378 m.s^{-1}$  

Similarly  

$v_2= -13.8- 9.8 \times 0.361$  ∵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:

$17.3378^{2}= 13.8^{2} + 2 \times 9.8 \times s$

$s= \frac{17.3378^{2}-13.8^{2}}{2 \times 9.8}$

$s = 5.62m$

For the body in the second case:

$(- 17.3378)^{2} = (- 13.8)^{2}- 2 \times 9.8\times s$

$s= \frac{300.6 - 190.44}{-2 \times 9.8}$

$s=- 5.62m$

∴The bodies are $2 \times 5.62m= 11.24m$ apart from each other from the point from which they are thrown.