Answer:
Explanation:
The motions of the two water ballons are ruled by the kinematic equations:
We are only interested in the vertical motion, so that equation is all what you need.
1. Water ballon is thrown horizontally at sped 2.00 m/s.
The time the ballon takes to hit the ground is independent of the horizontal speed.
Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.
Then:
[tex]y=y_0+V_0t-gt^2/2\\ \\ 0=6.00m-9.8\frac{m}{s^2} t^2/2\\ \\ t=\sqrt{2\times6.00m/9.8\frac{m}{s^2}}\\\\ t=1.11s[/tex]
2. Water ballon thrown straight down at 2.00 m/s
Now the initial vertical speed is 2.00 m/s down. So, the equation is:
[tex]0=6.00m-2.00\frac{m}{s}t-9.8\frac{m}{s^2}t^2/2\\ \\ 4.9t^2+2t-6=0\\ \\ t=0.92s[/tex]
To solve the equation you can use the quadratic formula.
[tex]t=\frac{-2+/-\sqrt{2^2-4(4.9)(-6)} }{2(4.9)}\\ \\ t=-1.33\\ \\ t=0.92[/tex]
You get two times. One of the times is negative, thus it does not have physical meaning.
3. Conclusion:
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
