Question

Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reached in their paths the samemaximum height. Ball A covered the total horizontal range D; ball B covered the total horizontal range 1.5D and ball C covered total horizontal range 2D when they hit the ground. Which of these three balls spent the greatest time in flight? (section 4.3) They all have the same time of flight ОА OB oc Band A and B

Answer 1

Answer:

• The time of the flight will be the same for the three balls.

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

$y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$

where $y_0$ is the initial position, $v_0$ the initial speed, and a the acceleration.

And the equation for the speed is:

$v(t) = v_0 + a t$

Now, lets say that the position is the height measured from the ground, then, in our problem, $y_0$ must be zero, $v_0$ is the y component of the velocity $v_{y_0}$ and a is the gravitational acceleration of the Earth, that points in the negative y direction:

$a = - g = - 9.8 \frac{m}{s^2}$.

Taking all this together, we get

$y(t) = \ v_{y_0} \ t \ - \frac{1}{2} \ g \ t^2$

and

$v(t) = v_{y_0} - g t$

Now, at the maximum height, the speed must be zero. I

$v(t_{maxh}) = 0 = v_{y_0} - g t_{maxh}$

but this means

$v_{y_0} = g t_{maxh}$

So, the initial y component of the velocity must be the same for the three balls.

Now, the equation for the position was

$y(t) = \ v_{y_0} \ t \ - \frac{1}{2} \ g \ t^2$

At time $t_{ground}$ the balls reach the ground, this is, height zero

$y(t_{ground}) = 0 = \ v_{y_0} \ t_{ground} \ - \frac{1}{2} \ g \ t_{ground}^2$

$0 = \ g t_{maxh} \ t_{ground} \ - \frac{1}{2} \ g \ t_{ground}^2$

$0 = \ g \ t_{ground} \ (t_{maxh} \- \frac{1}{2} \ \ t_{ground})$

so

$0 = (t_{maxh} \- \frac{1}{2} \ \ t_{ground})$

$t_{ground} = 2 t_{maxh)$

So, as we can see, the time of the flight will be the same for the three balls.