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## AP®︎/College Physics 1

# Impact velocity from given height

Determining how fast something will be traveling upon impact when it is released from a given height. Created by Sal Khan.

## Want to join the conversation?

- why not just use s=1/2at^2 solve for t than use V=at?(13 votes)
- I think that would be another way of doing it. Worked it out and it seems like you would get the same answer...(12 votes)

- Since the motion of the object and the acceleration (gravity) is in the same direction, aren't you supposed to take + 9.8 m/s^2 ?(7 votes)
- So how do you find the time? Just plug in the Vf in V/a?(0 votes)

- How is 9.8 m/sec^2 the affect gravity has on an object? Doesn't it start slower than that and start speeding up until it's faster than that?(3 votes)
- The velocity starts at 0 m/s, and then continues to increase as long as the object is falling. So, yes, the object does start slower and will eventually reach speeds faster than 9.8 m/s, but that is velocity and not acceleration. The acceleration remains constant, and is reflecting the increasing velocity of the object as it continues to fall. The object won't stop accelerating until it hits the ground.(4 votes)

- I am confused about the terminology used to describe acceleration and I need some clarification. If you are using a coordinate system where upwards is positive and you throw some thing upward is acceleration positive or negative? If upwards is positive then when you throw something up the acceleration is constantly increasing which to me seems incorrect. I was under the impression that if you throw something upwards the longer it's in the air the lower the acceleration until it starts falling back done and increasing again. Could someone please clarify this for me??(3 votes)
- If you label the up direction as positive and the force of gravity is down then the acceleration due to gravity is negative.

When you have up as positive and you throw an object up its initial velocity is positive and once it leaves your hand the acceleration of the object is only due to gravity and is constant at -9.8 m/s^2. This vertical acceleration doesn't change.

As this object travels its initial velocity will change by -9.8 m/s every second. Let's say the object was thrown up at 29.4 m/s. So since the object was thrown up which a positive direction it is initially traveling at + 29.4 m/s.

After 1 second we know that the velocity changed by - 9.8 m/s so at this point in time the object is traveling at a velocity of (+ 29.4 m/s) + (- 9.8 m/s) = + 19.6 m/s.

After another second, a total of 2 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be (+ 19.6 m/s) + (- 9.8 m/s) = + 9.8 m/s.

After another second, a total of 3 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be (+ 9.8 m/s) + (- 9.8 m/s) = 0 m/s.

After another second, a total of 4 seconds, the velocity will have changed by another - 9.8 m/s so that the velocity would be ( 0 m/s) + (- 9.8 m/s) = - 9.8 m/s.

And so on until the object lands.(5 votes)

- If the fall was infinite, could we reach an infinite velocity ?(2 votes)
- With the equations we're using, the answer is
**yes**-- if you had an infinite distance to fall, you would attain an infinite speed. This**doesn't**happen in real life, though, and the difference is because we are using equations which are slightly simplified. They are accurate enough for everyday use, but they give wrong answers for extreme cases such as "falling an infinite distance". The main differences are that (a) we assume the acceleration (due to gravity) will be constant, but this is only accurate if you're falling distances up to a few km; and (b) we assume that your acceleration doesn't depend on your current speed, but this is only accurate at "everyday" speeds (up to a few hundred km/s) -- in reality it takes more and more force to accelerate the faster you are already going, and it is impossible to accelerate to speeds faster than the speed of light, which is about 300,000 km/s.(8 votes)

- For the final question I got -35.64 kilometers per hour and 22.15 miles per hour.(4 votes)
- How much does air resistance affect the velocity or the acceleration(2 votes)
- Air resistance increases in proportion to the square of the speed of the object, so as the speed increases, the acceleration decreases. There comes a point where acceleration is zero; the object is now at terminal velocity. For a falling human (a sky-diver with closed parachute, for example) this is about 125 mph.(3 votes)

- Sal said we were trying to find the velocity before it hit the ground. Wouldn't the final velocity be when it actually hit the ground?(2 votes)
- Think of it as the velocity at the instant it touches the ground before it has actually started to come to rest.(2 votes)

- if you throw a rock off that, right at the bottom, right before it hits the ground, it will have a velocity of negative 9.9 meters per second. So negative 9.9 meters per second.

so if the velocity is any number that will be the ---- per sec?(2 votes)- The
**unit for velocity**is*m/s*. So, yes if the**velocity=any number**after that number there will be the unit*m/s*.(3 votes)

- I would like to know how do I go about calculating the angle using the formula R=(V^2sin2theta)/g with an unknown of velocity initial.

A ball is kicked off the ground reaching a maximum height of 60m and it lands 80m away from the starting point. calculate the initial velocity and the angle above the horizontal of the ball when it was kicked.

Could you also explain when do I use the different formulae of Range or horizontal distance(2 votes)

## Video transcript

What I want to do in this video
is answer an age-old question, or at least an interesting
question to me. And the question is, let's
say I have a ledge here-- I have a ledge or
cliff, or maybe this is a building of some kind. And let's say it has height, h. So let's say it has a height
of h, right over here. And what I'm curious about is
if I were to either-- Let's say that this is me over
here, so this is me. If I were to either
jump, myself-- that's not recommended
for very large h's. Or If I were to throw something,
maybe a rock off of this ledge, how fast would
either myself or that rock be going right before
it hits the ground? And like all of the
other videos we're doing on projectile
motion right now, we're going to ignore
air resistance. And for small h's and
for small velocities, that's actually reasonable. Or if the object
is very aerodynamic and is kind of dense,
then the air resistance will matter less. If it's me kind of belly
flopping from a high altitude, then the air resistance
will start to matter a lot. But for the sake of simplicity,
we're going to assume no air. Or we're not going
to take into effect the effects of air resistance. Or we could assume
that we're doing this on an Earth-like planet
that has no atmosphere. However you want to do it. So let's just think
about the problem. And just so you know,
some of you might say, that's not realistic. But this actually would be
realistic for a small h. If you were to jump off of the
roof of a one-story building, air resistance will not
be a major component in determining your speed if. It was to be a much larger
building, then all of a sudden it matters. And I don't recommend you
do any of these things. Those are all very
dangerous things. Much better to do
it with a rock. So that's actually the example
we're going to be considering. So let's just think
about this a little bit. We want to figure
out-- So at the top, right when the
thing gets dropped, right when the
rock gets dropped, you have an initial
velocity of 0. And once again, we're going
to use the convention here that positive velocity
means upwards, or a positive vector means up,
a negative vector means down. So we're going to have an
initial velocity over here of 0. And then at the
bottom we're going to have some final
velocity here that is going to be a
negative number. So it's going to have some
negative value over here. So this is going to be negative. This is going to be a negative
number right over there. And we know that the
acceleration of gravity for an object on free
fall, an object in free fall near the
surface of the earth. We know it, and we're going
to assume that it's constant. So our constant
acceleration is going to be negative 9.8 meters
per second squared. So what we're going
to do is given an h, and given that their
initial velocity is 0 and that our acceleration
is negative 9.8 meters per squared, we want to figure
out what our final velocity is going to be right before
we hit the ground. We're going to
assume that this h is given in meters,
right over here. And we'll get an
answer in meters per second for that
final velocity. So let's see how we
can figure it out. So we know some basic things. And the whole point of
these is to really show you that you can always derive
these more interesting questions from very basic
things that we know. So we know that displacement is
equal to average velocity times change in time. And we know that
average velocity-- if we assume
acceleration is constant, which we are doing--
average velocity is the final velocity plus
the initial velocity over 2. And then our change in time,
the amount of elapsed time that goes by-- this is
our change in velocity. So elapsed time
is the same thing. I write it over here--
is our change in velocity divided by our acceleration. And just to make sure
you understand this, it just comes
straight from the idea that acceleration-- or
let me write it this way-- that change in velocity is
just acceleration times time. Or I should say, acceleration
times change in time. So if you divide both sides of
this equation by acceleration, you get this right over here. So that is what our
displacement-- Remember, I want an expression
for displacement in terms of the things
we know and the one thing that we want to find out. Well, for this example
right over here, we know a couple of things. Well actually, let me
take it step by step. We know that our
initial velocity is 0. So this first expression
for the example we're doing, the average velocity is going
to be our final velocity divided by 2, since our
initial velocity is 0. Our change in velocity
is the same thing as final velocity
minus initial velocity. And once again, we know that
the initial velocity is 0 here. So our change in velocity
is the same thing as our final velocity. So once again,
this will be times. Instead of writing
change in velocity here, we could just write
our final velocity because we're starting at 0. Initial velocity is 0. So times our final velocity
divided by our acceleration. Final velocity is the same
thing as change in velocity because initial velocity was 0. And all of this is going
to be our displacement. And now it looks like
we have everything written in things we know. So if we multiply both
sides of this expression or both sides of this equation
by 2 times our acceleration on that side. And we multiply
the left-hand side by-- I'll do the same colors--
2 times our acceleration. On the left hand side, we
get 2 times our acceleration times our displacement
is going to be equal to, on the right hand side, the
2 cancels out with the 2, the acceleration cancels
out with the acceleration-- it will be equal
to the velocity, our final velocity squared. Final velocity times
final velocity. And so we can just solve
for final velocity here. So we know our acceleration
is negative 9.8 meters per second squared. So let me write this over here. So this is negative 9.8. So we have 2 times
negative 9.8-- let me just multiply that out. So that's negative 19.6
meters per second squared. And then what's our
displacement going to be? What's the displacement
over the course of dropping this rock off of
this ledge or off of this roof? So you might be tempted to say
that our displacement is h. But remember, these
are vector quantities, so you want to make sure
you get the direction right. From where the rock started to
where it ends, what's it doing? It's going to go
a distance of h, but it's going to go a
distance of h downwards. And our convention
is down is negative. So in this example,
our displacement from when it leaves your hand
to when it hits the ground, the displacement is going
to be equal to negative h. It's going to travel
a distance of h, but it's going to travel
that distance downwards. And that's why this vector
notion is very important here. Our convention is
very important here. So our displacement over here is
going to be negative h meters. So this is the
variable, and this is the shorthand for meters. So when you multiply
these two things out, lucky for us these
negatives cancel out, and you get 19.6h meters squared
per second squared is equal to our final velocity squared. And notice, when
you square something you lose the sign information. If our final velocity was
positive, you square it, you still get a positive value. If it was negative
and you square it, you still get a positive value. But remember, in
this example, we're going to be moving downward. So we want the negative
version of this. So to really figure
out our final velocity, we take, essentially,
the negative square root of both sides of this equation. So if we were to take the square
root of both sides of this, you take the square
root of that side, you take the square root of
that side, you will get-- and I'll flip them around-- your
final velocity, we could say, is equal to the
square root of 19.6h. And you can even take the
square root of the meter squared per second squared, treat
them almost like variables, even though they're units. And then outside of
the radical sign, you will get a
meters per second. And the thing I want
to be careful here is if we just take the
principal root here, the principal root here is
the positive square root. But we know that our velocity
is going to be downwards here, because that is our convention. So we want to make sure we
get the negative square root. So let's try it out
with some numbers. We've essentially
solved what we set out to solve at the
beginning of this video, how fast would we be falling,
as a function of the height. Well, let's try it
out with something. So let's say that the
height is-- I don't know, let's say the height
is 5 meters, which would be probably
jumping off of a or throwing a rock off
of a one-story, maybe a commercial one-story building. That's about 5 meters,
would be about 15 feet. So yeah, about the roof
of a commercial building, give or take. So let's turn it on. And so what do we get? If we put 5 meters in here, we
get 19.6 times 5 gives us 98. So almost 100. And then, we want to take
the square root of that, so it's going to be almost 10. So the square root of
98 gives us roughly 9.9. And we want the negative
square root of that. in that situation, when
the height is 5 meters-- So if you jump off of a
one-story commercial building, right at the bottom, or if you
throw a rock off that, right at the bottom, right
before it hits the ground, it will have a velocity of
negative 9.9 meters per second. So negative 9.9
meters per second. I'll leave it up to
you, as an exercise, to figure out how fast
this is either kilometers per hour or miles per hour
because it's pretty fast. It's not something
you's want to do. And this is just off of
a one-story building. But you can usually
figure this out. You could use this for,
really, any height as long as we're reasonably close
to the surface of the earth and you ignore the
effects of air resistance. At really high heights,
especially if the object is not that aerodynamic,
then air resistance will start to matter a lot.