# Stone being tossed from mountain Uni Physics with Calc

## Homework Statement

A stone is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45◦ . With what speed was it thrown?

## Homework Equations

g=9.8
projectile motion is symmetric

## The Attempt at a Solution

I got the answer 277.539 by treating this is a projectile motion problem where the stone is thrown upwards from the ground at a 45 degree angle (since the stone lands at a 45 degree angle it should depart the ground with the same angle since projectile motion is symmetric). I then made some equations for acceleration velocity and position in two dimensions. I solved for time when the y velocity was 0 (this is the midpoint/highpoint of the projectile path). This time value is in terms of the magnitude of the initial velocity. I plugged this time value in to the position in y equation and set it equal to 20 (20 is the height of the rock at the midpoint because the problem says that the rock is thrown off the rock horizontally implying that at this point there is no y velocity and that it is the high point of the path of motion). I solve for the magnitude of the initial velocity and get 40g+1/2. Then, I take this magnitude and multiply it by the cosine of 45 degrees since the components of a vector equals the magnitude times the cos(angle) sin(angle) and I use cos since I'm looking for the x component of the initial velocity. My final answer ends up being 277.5394 m/s. According to this practice test the correct answer is 20. I really don't get how what I did is incorrect. If someone could point out the flaw in my logic I'd be super thankful. Thanks!

DEvens
Gold Member
Symmetry there jon. It hits at 45 degrees. So the horizontal and vertical speeds are the same. What is the vertical speed?

Symmetry there jon. It hits at 45 degrees. So the horizontal and vertical speeds are the same. What is the vertical speed?
I got that the initial speed in the x and y direction is 277.5394, which is apparently wrong. Although the y speed is getting continuously lower since the acceleration is -g. X velocity stays constant at 277.5394.

DEvens
Gold Member
Ok, ask it a different way. A stone falls 20 meters starting at velocity zero. How fast is it going when it hits?

Ok, ask it a different way. A stone falls 20 meters starting at velocity zero. How fast is it going when it hits?
19.78 m/s

DEvens
SteamKing
Staff Emeritus
Homework Helper
I got that the initial speed in the x and y direction is 277.5394, which is apparently wrong. Although the y speed is getting continuously lower since the acceleration is -g. X velocity stays constant at 277.5394.

You should realize that 277 m/s is about how fast a jet airliner is traveling when it reaches full speed. oo)

You should realize that 277 m/s is about how fast a jet airliner is traveling when it reaches full speed. oo)
Yeah I know it's not right. I just don't get what I did wrong.

19.78 m/s
So I get why this is the answer, but I really want to know why what I did didn't work. I'm going to upload a picture of my work and maybe somebody can point out the error.

SteamKing
Staff Emeritus
Homework Helper
Yeah I know it's not right. I just don't get what I did wrong.
That's why we ask users of the HW forums to post all their calculations. It saves time and second guessing on the part of those choosing to respond.

haruspex
Homework Helper
Gold Member
So I get why this is the answer, but I really want to know why what I did didn't work. I'm going to upload a picture of my work and maybe somebody can point out the error.
Never mind about the diagram, your described method is clear. Just post your working (NOT as a picture).

## Homework Statement

A stone is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45◦ . With what speed was it thrown?

## Homework Equations

g=9.8
projectile motion is symmetric

## The Attempt at a Solution

I got the answer 277.539 by treating this is a projectile motion problem where the stone is thrown upwards from the ground at a 45 degree angle (since the stone lands at a 45 degree angle it should depart the ground with the same angle since projectile motion is symmetric). I then made some equations for acceleration velocity and position in two dimensions. I solved for time when the y velocity was 0 (this is the midpoint/highpoint of the projectile path). This time value is in terms of the magnitude of the initial velocity. I plugged this time value in to the position in y equation and set it equal to 20 (20 is the height of the rock at the midpoint because the problem says that the rock is thrown off the rock horizontally implying that at this point there is no y velocity and that it is the high point of the path of motion). I solve for the magnitude of the initial velocity and get 40g+1/2. Then, I take this magnitude and multiply it by the cosine of 45 degrees since the components of a vector equals the magnitude times the cos(angle) sin(angle) and I use cos since I'm looking for the x component of the initial velocity. My final answer ends up being 277.5394 m/s. According to this practice test the correct answer is 20. I really don't get how what I did is incorrect. If someone could point out the flaw in my logic I'd be super thankful. Thanks!

Here are pictures of my work:http://i.imgur.com/JwVipFV.jpg
http://imgur.com/qkqyf6h

Never mind about the diagram, your described method is clear. Just post your working (NOT as a picture).
Okay so I'll try to write out everything I did here with slashes and ^. So accel. equals <0,-g> then I integrate to get vel. = <sqrt2/2*magnitude of the initial velocity,sqrt2/2*magnitude of the initial velocity - gt> integrate again to get pos = <(sqrt2*t)/2*magnitude of the initial velocity, (sqrt2*t)/2*magnitude of the initial velocity - g/2t^2> I then set the y velocity equal to zero. This gets me t = sqrt2/(2g). Then I plug this t value into the y position and set it equal to 20, then I solve for the magnitude of the initial velocity. Then I multiply the magnitude of the initial velocity by sqrt2/2 to get the velocity in the x direction.

SteamKing
Staff Emeritus
Homework Helper
It's not clear where your motion equations come from. They don't appear to be the normal SUVAT equations. I don't follow your integrations either.
Accel. in x is 0 accel. in y is -g.
vel. in x is whatever the initial velocity in the x direction was. vel. in y is -gt + whatever the initial velocity in the y direction was.
pos in x is whatever the initial velocity in the x direction was times t. pos in y is -g/2t^2 + t times whatever the initial velocity in the y direction was

the initial velocity in the x and y direction is both cos(45)*magnitude of the initial velocity

make some sense now?

SteamKing
Staff Emeritus
Homework Helper
Accel. in x is 0 accel. in y is -g.
vel. in x is whatever the initial velocity in the x direction was. vel. in y is -gt + whatever the initial velocity in the y direction was.
pos in x is whatever the initial velocity in the x direction was times t. pos in y is -g/2t^2 + t times whatever the initial velocity in the y direction was

the initial velocity in the x and y direction is both cos(45)*magnitude of the initial velocity

make some sense now?

No, the equations you have written in your calculations still don't make any sense.

If you take your trajectory as sketched, the rock starts at some initial velocity v0 at an angle of 45°. By the time the rock has reached an altitude of 20 m, the vertical component of its velocity must equal 0. This implies that to fall back the 20 m to the ground will take an equal amount of time as it took to get to the top of the hill,

or h = 20 m = (1/2) * g * t2.

When the rock reaches an altitude of 20 m, the vertical velocity = 0, which means the initial velocity must be v0 * sin (45°) - g * t = 0.

To find v0, solve for t using the first equation, and then plug this value of t into the second equation. Once you have calculated v0, you can calculate the horizontal velocity required to throw the rock from the top of the hill and have it strike the ground at a 45° angle.

I can't tell where your reasoning went awry, but it turned the solution to this problem into something way more complicated than it should have.