Answer:
Explanation:
This is very interesting problem. The key to solve it is to understand that we don't care about the horizontal movement of the balls. Let's see why using kinematics:
We know, that, in 1D, for constant acceleration, the position y at time t is given by
[tex]y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2[/tex]
where [tex]y_0[/tex] is the initial position, [tex]v_0[/tex] the initial speed, and a the acceleration.
And the equation for the speed is:
[tex]v(t) = v_0 + a t[/tex]
Now, lets say that the position is the height measured from the ground, then, in our problem, [tex]y_0[/tex] must be zero, [tex]v_0[/tex] is the y component of the velocity [tex]v_{y_0}[/tex] and a is the gravitational acceleration of the Earth, that points in the negative y direction:
[tex]a = - g = - 9.8 \frac{m}{s^2}[/tex].
Taking all this together, we get
[tex]y(t) = \ v_{y_0} \ t \ - \frac{1}{2} \ g \ t^2[/tex]
and
[tex]v(t) = v_{y_0} - g t[/tex]
Now, at the maximum height, the speed must be zero. I
[tex]v(t_{maxh}) = 0 = v_{y_0} - g t_{maxh}[/tex]
but this means
[tex] v_{y_0} = g t_{maxh}[/tex]
So, the initial y component of the velocity must be the same for the three balls.
Now, the equation for the position was
[tex]y(t) = \ v_{y_0} \ t \ - \frac{1}{2} \ g \ t^2[/tex]
At time [tex]t_{ground}[/tex] the balls reach the ground, this is, height zero
[tex]y(t_{ground}) = 0 = \ v_{y_0} \ t_{ground} \ - \frac{1}{2} \ g \ t_{ground}^2[/tex]
[tex] 0 = \ g t_{maxh} \ t_{ground} \ - \frac{1}{2} \ g \ t_{ground}^2[/tex]
[tex] 0 = \ g \ t_{ground} \ (t_{maxh} \- \frac{1}{2} \ \ t_{ground})[/tex]
so
[tex] 0 = (t_{maxh} \- \frac{1}{2} \ \ t_{ground})[/tex]
[tex] t_{ground} = 2 t_{maxh)[/tex]
So, as we can see, the time of the flight will be the same for the three balls.
