Respuesta :
Answer:
11.35 m/s
Explanation:
Let the first ball takes time t to reach the ground.
Use second equation of motion
[tex]s=ut + 0.5at^{2}[/tex]
Here, s = 15 m, a = 9.8 m/s^2, u = 0 m/s
By substituting the values
15 = 0 + 0.5 x 9.8 x t^2
t = 1.75 second
Now time taken by the first ball to reach 3.20 m is t'
Use second equation of motion
[tex]s=ut + 0.5at^{2}[/tex]
Here, s = 3.2 m, a = 9.8 m/s^2, u = 0 m/s
By substituting the values
3.2 = 0 + 0.5 x 9.8 x t'^2
t' = 0.81 s
Thus, the time for the second ball to reach the ground is t - t' = 1.75 - 0.81 = 0.94 s
For second ball:
Use second equation of motion
[tex]s=ut + 0.5at^{2}[/tex]
Here, s = 15 m, a = 9.8 m/s^2, u = ?, t = 0.94 s
By substituting the values
15 = u x 0.94 + 0.5 x 9.8 x 0.94 x 0.94
u = 11.35 m/s
Thus, the second ball has the initial velocity of 11.35 m/s.
Answer:
[tex]v = 11.34 m/s[/tex]
Explanation:
After 3.20 m of free fall the time taken by the stone is given as
[tex]d = \frac{1}{2}gt^2[/tex]
so here we will have
[tex]3.20 = \frac{1}{2}(9.8)(t^2)[/tex]
[tex]t = 0.81 s[/tex]
so speed of the stone is given as
[tex]v_f = 0 + 9.81(0.81)[/tex]
[tex]v_f = 7.93 m/s[/tex]
now when other stone is thrown downwards
then the relative speed of the two stones is given as
[tex]v_r = v_1 - v_2[/tex]
[tex]v_r = v - 7.93[/tex]
now the time taken by the stone to drop down by 15 m
[tex]15 = \frac{1}{2}(9.81) t^2[/tex]
[tex]t = 1.75 s[/tex]
so the time remaining to catch the two stones is given as
[tex]\Delta t = 1.75 - 0.81 = 0.94 s[/tex]
so we have
[tex](v - 7.93)\times 0.94 = 3.20[/tex]
[tex]v = 11.34 m/s[/tex]
