# Projectile motion stone throw problem

vysero
A stone is projected at a cliff of height h with an initial speed of 42m/s directed at angle=60 above the horizontal. The stone strikes at A, 5.5s after launching. Find A) the height h of the cliff, B) the speed of the stone just before impact at A, and C) the maximum height H reached above the ground.

ok so,

h= Vo(sin(60))t-1/2g(t)^2

so i have my answer for a which came out to be 51.8m

Here is where I am struggling with problem A. I only understand a small portion of why: h= Vo(sin(60))t-1/2g(t)^2

For instance why is Vo = Vo(sin(60)) ok so sin is opp/hyp the angle is 60 that gives me the length of opposite side so why is that not the answer for C?

I am obviously completely confused.... help me...

## Answers and Replies

azizlwl
The easiest way to understand projectile is to draw a vertical component graph of velocity vs. time.
The horizontal component velocity remains constant.

The area under the curve is the total displacement.
From the drawing you will see negative area and negative area.

bfusco
A stone is projected at a cliff of height h with an initial speed of 42m/s directed at angle=60 above the horizontal. The stone strikes at A, 5.5s after launching. Find A) the height h of the cliff, B) the speed of the stone just before impact at A, and C) the maximum height H reached above the ground.

ok so,

h= Vo(sin(60))t-1/2g(t)^2

so i have my answer for a which came out to be 51.8m

Here is where I am struggling with problem A. I only understand a small portion of why: h= Vo(sin(60))t-1/2g(t)^2

For instance why is Vo = Vo(sin(60)) ok so sin is opp/hyp the angle is 60 that gives me the length of opposite side so why is that not the answer for C?

I am obviously completely confused.... help me...

well you understand at least the the initial velocity given is whats called a vector, meaning it has components (separate velocities in the x and y directions). Also a side note, it may be easier to mentally visualize these problems if you break things down and look at things in the x direction separately from the y direction.

now the equation you are using (and most of the other equations you probably have) h=Vot+.5at^2, has to be dealt with completely in the y direction. for Vo you cant just use the value of 42 m/s given because thats a combination of the x velocity and the y velocity. so the initial y velocity = Vosin60, so that goes into the equation as your Vo, t is independent of x and y direction of travel so that is easy 5.5 sec. the acceleration in that problem is obviously g which you got.

the answer you got doesn't work for part C because if you think about it, if you were to launch a rock of a cliff yourself and you were launching it not directly horizontally (meaning there is an angle) the rock would travel up before it begins traveling down. so the answer to part C would be the amount the rock traveled up + the height of the cliff. now as a clue as to figure out how far up the rock travelled, think about it completely in the vertical (y direction). The rock travels up and then what eventually happens to the velocity in the y direction?

vysero
So the cos of 60 is .866. So then .866(42) is equal to the Vo in the y direction? So then cos of 60 is 1/2 does that mean that 42(1/2) is my Vo in the x direction?

Second question, can I consider acceleration as a velocity in the x direction + a velocity in the y direction?

"so the answer to part C would be the amount the rock traveled up + the height of the cliff" what does the height of the cliff have to do with the amount the rock travels vertically? I mean say the cliff weren't there, it would still travel the same height up wouldn't it I am sure that's not what you were saying so I guess im confused to what your trying to tell me.

bfusco
So the cos of 60 is .866. So then .866(42) is equal to the Vo in the y direction? So then cos of 60 is 1/2 does that mean that 42(1/2) is my Vo in the x direction?

Second question, can I consider acceleration as a velocity in the x direction + a velocity in the y direction?

"so the answer to part C would be the amount the rock traveled up + the height of the cliff" what does the height of the cliff have to do with the amount the rock travels vertically? I mean say the cliff weren't there, it would still travel the same height up wouldn't it I am sure that's not what you were saying so I guess im confused to what your trying to tell me.

for your first question you have your sin and cos backwards. Vo in the y direction is sin 60. if you draw your V initial line at an angle of 60 degrees on a graph, you will see that the y component is opposite the angle thus sin60, and the x comp is adjacent the angle thus cos 60.

acceleration is never considered a velocity, acceleration is the rate at which velocity changes NEVER CONFUSE THE TWO lol. what i think you mean is does it also have components like the velocity did? and the answer would be yes it does, but for projectile motion there isnt an acceleration in the xdirection, it is always 0 at the end of the problem if you were asked the velocity in the x direction just before the rock hit the ground it would be the exact same as the velocity in the x direction of when it was launch IT NEVER CHANGES!!! <--very important. acceleration in the y direction (for projectile motion) is always gravity 9.8 m/s^2.

if you were standing at the edge of a cliff 30 meters high and you through the rock straight up and it reached a height of 10 meters above your head, then began falling, but over the edge of the cliff the rock would continue falling until it reaches the ground below the cliff. so that total distance traveled would be the height of the cliff 30 meters + the 10 meters you threw the rock above your head = 40 meters, your part C is asking the max height so in this example the max height would have been 40 meters.

just one final important note about my example, (keep in mind this is completely in the y direction) when you threw the rock straight up it had an initial velocity, but eventually slowed down until it reached a velocity 0, then it turns around and falls back to the earth.

so you can solve for the height above your head in your problem by using Vf^2=Vo^2+2aΔy, your final velocity would be 0 because your solving at the point when the rock stops traveling up and begins traveling down, and your initial velocity would be your Vosin60 and you just solve for Δy. that Δy + the total height of the cliff = your max height. Hope that helps

azizlwl
A stone is projected at a cliff of height h with an initial speed of 42m/s directed at angle=60 above the horizontal. The stone strikes at A, 5.5s after launching. Find A) the height h of the cliff, B) the speed of the stone just before impact at A, and C) the maximum height H reached above the ground.

ok so,

h= Vo(sin(60))t-1/2g(t)^2

so i have my answer for a which came out to be 51.8m

Here is where I am struggling with problem A. I only understand a small portion of why: h= Vo(sin(60))t-1/2g(t)^2

For instance why is Vo = Vo(sin(60)) ok so sin is opp/hyp the angle is 60 that gives me the length of opposite side so why is that not the answer for C?

I am obviously completely confused.... help me...

I assume point A is at the base of the cliff.

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vysero
Sorry there must have been some confusion with the question. The question is saying I am on flat ground and throwing a rock (so to say) on top of a cliff at an angle of 60. It takes the rock 5.5sec to hit point A which is on top of the cliff little h is the height of the cliff and big H is the apex of the arc of the rock or when velocity is 0 again.

Homework Helper
You can find vy buy using
Vy= vosinθ - gt and
Vx = vocosθ.
From these two components you can find the velocity at A.
You know the vertical component of the velocity. From that you can find H.

azizlwl
For instance why is Vo = Vo(sin(60)) ok so sin is opp/hyp the angle is 60 that gives me the length of opposite side so why is that not the answer for C?
...............................
Sorry i didn't read it carefully.

1. It's not Vo = Vo(sin(60)) but Vy=v0Sin(60°) and Vx=v0Cos(60°)

When the stone started flying, gravity will exert a force on the stone thus changing it's velocity. Newton's 2nd Law.
In horizontal direction, ignoring air resistance, the velocity remains constant, Newton 1st. Law.

Thus we have two separate velocities thus 2 equations in x and y direction that totally independent of each other.

2. The height is totally on y direction. Nothing to do with x direction. The only common between them is the time since obviously it is a single object, exist at same location(different value in x postion and y postion) at the same time.

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vysero
Ok let me rephrase my question:

Y-Yo = VoY(t)-1/2(g)(t)^2

This equation will give me the height of the cliff... why?

Y-Yo is the distance between Y final and Y initial how is that equal to the height of the cliff? If anything intuitively you would think this equation would somehow give me my maximum height.

Let me try and break the equation down:

VoY = velocity in the Y direction = Vo(sin(60))
t = 5.5 sec (why do I multiply my velocity in the y direction by 5.5 sec?)
-1/2(g) = -1/2(9.8) = -4.9 (why are we taking half of gravity and since we are why isn't it written like this -4.9???)
t^2 = 30.25 ( why am i multiplying by time x time, i thought we already included time???)

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Homework Helper
In part C you have to find H where vertical component of the velocity is zero.