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This question only exists in one dimension, the y axis.

A ball is thrown upwards with an initial velocity $V_o$ from a roof with height $h$. One second later, a second ball is dropped from the same roof. If $V_o$ of the first ball is greater than some value $V_{max}$,a value of $h$ does not exist that allows both balls to hit the ground at the same fime. Solve for $v_{max}$

This is my attempt at the solution, note: it is wrong.

They will meet at value $\Delta h_1 = \Delta h_2$

$\Delta h_1 = V_o*t + \frac{1}{2}*a*t^2$

$\Delta h_2 = \frac{1}{2}*a*(t-1)^2$

Let $n = \frac{1}{2}a$

$\Delta h_1 = V_o*t + n*t^2$

$\Delta h_2 = n*(t-1)^2$

$V_o*t + n*t^2 = n*(t-1)^2$

When I simplify further, I now have

$(V_o + 2n)*t = n$

This expression will give me the time it takes for the two values to be equal to each other.

So now if I substitute for t to get the time it takes the second ball to hit the ground,

$t = \sqrt{\frac{\Delta h}{n}} + 1$

I now get

$(V_o + 2n)*(\sqrt{\frac{\Delta h}{n}} + 1) = n$

So an impossible height for $h$ is anything greater than or equal to 0. If I set it to 0,

$V_0 = -n$

Solving this, I get

$V_0 = 4.9 \frac{m}{s}$

However, this is incorrect. The answer is supposed to be $9.8 \frac{m}{s}$

How am I supposed to do this question?

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closed as off-topic by ACuriousMind, AccidentalFourierTransform, user36790, Kyle Kanos, Gert Apr 15 '16 at 1:41

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  • $\begingroup$ That's the minimum velocity. The maximum is going to involve a very large $|\Delta h|$. $\endgroup$ – Bill N Apr 14 '16 at 21:13
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I will provide an intuitive way to solve this question.

Basically, if the first ball has not begun it's descent when the second ball is dropped, there is no way they can hit the ground simultaneously regardless of h. This is apparent if you think deeply about the situation. Imagine the first ball has not begun it's descent when the second ball is dropped. In this case, the second ball is closer to the ground and it has a greater downward velocity, so it is inevitable that the second ball will always hit before the first ball in this case. If however, if the first ball has already begun its descent after 1 second when the second ball is dropped, even though the second ball has a shorter distance to travel, the first ball will have a greater downward speed at t = 1, and therefore a height can be chosen so as to allow the first ball to catch up and to thus have the balls hit the ground simultaneously.

Mathematically we therefore want the first ball to have at least reached it's peak height at t = 1.

Using the formula,

$v = u + at$

The initial velocity $u = V_{max}$ , the final velocity at peak height $u = 0$ at $t = 1$ gives,

$0 = V_{max} + a$, and since $a = -9.8m/s/s$ we obtain,

$0 = V_{max} + -9.8$, therefore,

$V{max} = 9.8$

This means that if the first ball is thrown with an initial velocity of 9.8m/s, the ball will peak after 1 second. If the ball is thrown with a greater velocity, the ball will not have reached it's peak when the second ball is dropped and therefore will never be able to catch up to the second ball if the initial velocity of the first ball is greater than 9.8m/s.

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