Homework: Expression for minimum speed to cause collision of objects / simple kinematics

Question

I have a question which is worded as follows:

Person A throws a rock at a 45∘ angle to the horizontal, aiming at Person B who's standing on a building at heigh h above level ground. At the instant Person A throws their rock, Person B drops another rock. The two rocks will collide, no matter what Person A's rock's initial speed, provided it's greater than some minimum value. Find an expression for that minimum speed.

So far we have only been introduced to the following equations and calculus is not covered in this course.

v = u + at
s = ut + 1/2at²
v² = u² + 2as

I'm at a loss as to how to approach this problem. The answer must be an equation given in terms of v.

My thought process was to vectorise the x and y components of the baseball being thrown and compare to the other ball dropping, but got lost down that path.

If someone could please show me the correct approach with a problem like this it would be appreciated. I'm not looking to be spoon fed the answer, I want to learn how to correctly tackle problems like this as I know something like this will be on the exam.
Thanks

Answer 1

Let's first examine the position of Person B's rock. This rock will have an initial velocity of 0 and will accelerate downwards at $g=9.8m/s/s$.

Applying formula for displacement, $s = ut + 0.5at^2$, we arrive at,

$s = -0.5gt^2$

The position of the rock at any time is thus given by,

$s_{b_y} = h - 0.5gt^2$ (since the initial position is h)

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Now let's examine the position of rock A at any time.

Let $v$ be the initial velocity of this rock.

Thus in the x-direction (where the is no acceleration), we can use the formula $v\cos 45 = \frac{s_x}{t}$.

Therefore, $s_x = vt \cos 45$

Now let's consider the y-direction, by applying the formula $s = ut + 0.5at^2$,

$s_{y} = vt\sin 45 - 0.5gt^2$

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Now that we have the position of each rock at any time, we can see what is required for the rocks to hit (that is positions and times match for the two rocks).

Firstly we require the horizontal position of rock a to be h units from the starting point. That is,

$s_x = vt \cos 45 = h$

Re-arranging gives,

$t = \frac{h}{v\cos 45}$

Thus the interaction must occur at the time specified by the above equation. At this time, we will require the vertical position of each rock to match. That is,

$s_y = s_{b_y}$,

$vt\sin 45 - 0.5gt^2= h -0.5gt^2$

That is, $vt\sin 45 = h$

Utilising the fact that $t = \frac{h}{v\cos 45}$, we have,

$\frac{vh\sin 45}{v\cos 45} = h$

Which for any v>0 gives,

$h = h$

Therefore, it appears at first glance that for all velocities the two will meet.

But this is wrong in reality because this mathematical treatment assumes the particles can sometimes meet below the ground.

The minimum velocity will thus be the minimum velocity required for rock A to travel a distance of h before reaching the ground.

To find this velocity, we first need to find the time the rock hits the ground, that is when %s_x = 0%. That is,

$0 = v\sin 45 t - 0.5gt^2$,

$0 = t(v \sin 45 - 0.5gt)$

Thus,

$t = \frac{v \sin45}{0.5g}$

Now at this time, $s_y = h$,

Again in the horizontal direction, velocity is constant, thus

$v\cos 45 = h/t$

Plugging in the above time gives,

$v\cos 45 = \frac{h}{\frac{v \sin45}{0.5g}}$

Re-arraigning (and making use of the fact $sin 45 = cos 45 = \frac{1}{\sqrt{2}}$ gives,

$v = \sqrt{gh}$