projectile problem

Question

a stone is thrown vertically upward. given elevation of 50 m and initial velocity of 15 m/s

 

find:

 

a)maximum height

b) time it took to reach maximum height

c)velocity before it hits ground

d)time total stone is in air

 

 

need help pls

Answer 1

Hi Pia!

Let's put this problem in some context, so you can also solve others like it.

Kinematics, the subject of this problem, is the study of motion using five key quantities: acceleration (a), initial velocity (vo), final velocity (v), displacement (x-x_o, where x is final position and x_o is initial position), and time (t). Usually, a problem asks you to solve for one of these quantities. You should have four equations at your disposal, relating these five quantities. Each equation involves four of the five quantities. Therefore, if you want to use the equations to solve for one of the kinematic quantities, you have to know at least three others for the given situation. The challenge of kinematic problems is usually finding out what information you are given.

 

(though this is listed as a projectile problem, it is clearly only in one dimension, the vertical, so we do not need to choose which dimension to work in first)

The way I usually work these problems is to ask, first, which of the five kinematic quantities is being asked for? This is often easiest to pick out, because the question has to directly state it.  In part a), it is said that you are to find maximum height.  This means solving for displacement.  So:

 

to find:  (x-x_o)

 

Now, you have to figure out which three kinematic quantities are known.

 

A projectile, by definition, is an object moving under the influence of only gravitational acceleration.  So, we always know a projectile only has one acceleration on it, in the vertical, and it is a = g, down.  I usually call down negative, and -- though you do not have to -- it is so common to call down negative that I think it is good practice.  So, a = -g = -9.8 m/s².  That's one kinematic quantity we know.

 

We are told directly that its initial velocity is 15 m/s, upward.  So v_o = (+)15 m/s.  That's two.

 

Finally, we have to know a third kinematic quantity.  Since we are solving for (x-x_o), and a and v_o are already taken, our remaining possibilities are v and t.  We can look down in the problem and see that t is meant to be solved for later, so it is probably not given or known. 

 

That leaves v, the final velocity.  Though it is not directly stated, any time you see that you are dealing with a circumstance that ends with the "maximum height" for a projectile, you can be sure you are dealing with a case where its final velocity over that interval is 0.  If a projectile has a non-zero vertical velocity, it is either still going toward its maximum height, or coming back down from its maximum height.  So v = 0.  That's three.

 

So we know:

 

known:  a (= -g), v_o (= 15 m/s), v (=0)

 

Now that you have the three known quantities, take a look at your stable of kinematic equations, and see which one involves the three things you know, and the one you are looking for:  (x-x_o), v_o, v, and a.  As long as one of these four quantities is initial velocity, you will always have one of your equations that contains only these four.  In this case, it would be:

 

v² = v_o²+2a(x-x_o)

 

Now, we drop in our values and solve for what we want to find:

 

0 = (15 m/s)²+2(-9.8 m/s²)(x-x_o) --> 0 = 15²-2(9.8)(x-x_o)

 

Solving this for (x-x_o) will give you the final displacement, which is the height (x) that the projectile reaches above the starting height (x_o = 50 m).  So you can immediately solve for x upon knowing x-x_o.

 

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Part b is made easier by Part a, since all the other kinematic quantities over this interval, besides time, are now known.  So you have your full suite of kinematic equations available to you, and you can use any one that involves time.  I prefer to use the one which requires the least algebra, which would be:

 

v = v_o + at  (this has the added advantage, in this case, of only involving quantities we are given, so any errors we might make in Part a do not propagate down to this part)

 

Putting in what we know:

 

0 = (15 m/s)+(-9.8 m/s²)t --> 0 = 15-9.8t

 

which can be solved for t.

 

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For Parts c and d, we reset the interval of interest to be the entire time from launch to when it hits the ground, 50 m below the launch height.  We have to refigure what we want to find, and what we know, over this new interval.

 

Starting with what we want to find, which is again directly stated:  final velocity.

 

to find: v

 

We are still dealing with a projectile, so we still known a = -g = -9.8 m/s².  We are also still dealing with the same launch.  so we still have v_o = 15 m/s.

 

But, over this interval, we are solving for v, so we know longer know it (as we did in Parts a and b).  So, we now have to know one more kinematic quantity from among the two remaining for this interval:  (x-x_o) and t.  We are solving for t in the next part, so we can be reasonably sure it is not yet known.

 

That leaves (x-x_o).  We know the projectile ends up 50 m below where is starts, so x = -50 m.  This means (x-x_o) = (-50 m - 0 m) = - 50 m.

 

So now we have our three knowns that we need:

 

known: a (=-9.8 m/s²), v_o (=15 m/s), (x-x_o) (= -50 m)

 

Now we choose the kinematic equation involving (x-x_o), v_o, v, and a.  It is the same one as for Part a:

 

v² = v_o²+2a(x-x_o)

 

Putting in our known values:

 

v² = (15 m/s)² +2(-9.8 m/s²)(-50 m)

 

One wrinkle about using this particular kinematic equation to solve for either v or v_o is that they are both squared terms.  When we take the square root of a number, n, we are used to automatically taking the positive square root (the principal square root, as it is called in algebra).  HOWEVER, there is also always a negative square root (for example, the square root of 9 is either +3 OR -3).  

 

So, when using this equation to solve for v (or v_o), you have to have knowledge of the sign of that velocity from other information.  In this case, we can be sure that, when the projectile is about to hit the ground, its velocity is down.  Hence, you need to take the negative square root given by the previous equation as the solution for v.

 

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For Part d, we once again have a case where (due to our work in Part c) all the kinematic quantities are known for the given interval, so we just have to choose one that lets us solve for time.  Once again, I choose the one that involves the least algebra.  That would be:

 

(x-x_o) = (1/2)(v_o+v)t

 

Using the value for v determined in Part c (remember to keep that negative sign!  direction is important in kinematic equations), you can set up this equation ass:

 

-50 m = (1/2)(15 m/s + (v from Part c))t

 

and solve for t.

 

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This technique of determining what you want to find and the three things you know will work for just about all kinematics problems (there are a few wrinkles sometimes, but this general technique is widely applicable).  For two-dimensional projectile motion, you also have the added step of choosing which direction (horizontal or vertical) to work in first, but that is for another time.

 

Hope this helps!  If you want to check any answers, or have any more questions, just let me know.

 

Answer 2

coord sys is up, and the ground is y=0, a = -g.
a) v^2 - v0^2 = -2g(y-y0)
v=0,y0=50m. Solve for y.
b)v=v0-gt
at max height, v=0. v0=15. Solve for t.
c) at 2Xt in b, v=15m/s down at 50m.
v^2-v0^2 = -2g(y-y0)
v0=15, y=0, y0=50, solve for v
coord sys down here
d) 2Xt found in b +
t found with
v=v0+gt
where v=v in c) and v0 = 15.
(here use v, v0, and g are all down).