- #1
ConfusedMonkey
- 42
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Homework Statement
A ball is dropped from rest at height ##h_0## above the ground. At the same instant, a second ball is launched with speed ##v_0## straight up from the ground, at a point directly below where the other ball is dropped. (a) Find a condition on ##v_0## such that the two balls will collide mid-air. (b) Find an expression for the height at which they collide.
Homework Equations
The Attempt at a Solution
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Attempted solution for (a): I think all that is required is that ##v_0 > 0##. As long as the velocity is positive then the ball that is launched upward will be in the air by the time it collides with the ball that was dropped. But the textbook gives as a solution that we must have ##v_0 > \sqrt{gh_0/2}## where ##g = 9.8 m/s^2##. What's wrong with my thinking?
Attempted solution for (b): Let ##x_{dropped}(t)## be the position function of the ball that's dropped. Then ##x_{dropped}(t) = \frac{g}{2}t^2 + 0(t) + h_0 = \frac{g}{2}t^2 + h_0##. Now let ##x_{launched}(t)## be the position function of the ball that is launched. Then ##x_{launched}(t) = \frac{g}{2}t^2 + v_0t + 0 = \frac{g}{2}t^2 + v_0t##.
I can find at what time the two balls collide by equating the two above equations and solving for ##t##:
##x_{dropped} = x_{launched}##
##h_0 = v_0t##,
which gives ##t = \frac{h_0}{v_0}##. Plugging this solution into either ##x_{dropped}## or ##x_{launched}## gives the height of the collision as ##\frac{g}{2}\big(\frac{h_0}{v_0}\big)^2##. But the textbook's solution is ##h_0 - \frac{gh_0^2}{2v_0}##.
Could someone please tell me why my answers are wrong? Thanks.