# Projectile Motion - Rock Thrown Horizontally Off a Cliff

## Homework Statement

A student stands on the edge of a cliff and throws a stone horizontally over the edge with a speed "v1". The cliff is "h" meters high. Given [h,v1], Determine:
a. The time to hit the ground
b. The horizontal distance traveled
c. The magnitude and direction of the stone's velocity just before hitting the ground

## Homework Equations

x direction
$$V_x=V_{0x}$$
$$x=V_0\cos{\theta}$$

y direction
$$V_y=V_{0y}-gt$$
$$y=V_{0y}t-\frac{1}{2}gt^2$$
$$V_y^2=V_{0y}^2-2gy$$

## The Attempt at a Solution

If I set up my coordinate system so that 0 is ground level, and h is cliff level (where the stone was thrown from), and the distance to the landing point of the stone when it hits the ground is, say "r" meters from 0 in the x direction.

I've shown what work I have in the attached images, but to be perfectly honest, I'm having trouble with the entire basic strategy to approaching this problem.

When I try to solve for x distance r, I need time. So when I try to solve for time, that becomes reliant on r. I can't even think of anything else I can find.

I feel I'm either not utilizing the possible angles within this problem, or I've completely missed some fundamental idea in regards to projectile motion.

A good shove in the right direction on this kind of problem would be VERY much appreciated.

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Here's a simple approach:-

Firstly make a table of Horizontal motion and vertical motion, like according to your question,

H-Horizontal motion
V-Vertical motion
u-initial velocity
s-displacement
h-Height of the cliff
x and t are assumed since we have to find them out.

Ok, first lets see in which column, assumed variable are less. Its obvious that in V column, there is only one variable "t". So can you apply equation of motion here?

Something tells me I've once again forgotten a rather important fact about the equation
$$y=V_{0y}t+\frac{at^2}{2}$$

Is it correctly written as this by chance :grumpy:
$$(y-y_0)=V_{0y}t+\frac{at^2}{2}$$

Something tells me I've once again forgotten a rather important fact about the equation
$$y=V_{0y}t+\frac{at^2}{2}$$

Is it correctly written as this by chance :grumpy:
$$(y-y_0)=V_{0y}t+\frac{at^2}{2}$$
You can take y0 as 0 since the initial displacement is zero.

Now just plug in the values and you will get the time. :)

$$(y-y_0)=V_{0y}t+\frac{at^2}{2}$$
$$(0-h)=0-\frac{gt^2}{2}$$
$$2h=gt^2$$
$$\frac{2h}{g}=t^2$$
$$\sqrt{\frac{2h}{g}}=t$$

That's right.

Now just see that we have t same for both H and V motion. So now we are left with x in H motion. Try applying the suitable equation of motion in the H column.

Here is my attempt for Horizontal distance covered:
$$x=(V_0\cos{\theta})t$$
$$x=V_1t$$
$$x=\frac{V_1\sqrt{2h}}{g}$$
hows that look?

Now in order to get the direction and magnitude of the end velocity of the rock, I need both the Vx and Vy right?

Here is my attempt for Horizontal distance covered:
$$x=(V_0\cos{\theta})t$$
$$x=V_1t$$
$$x=\frac{V_1\sqrt{2h}}{g}$$
hows that look?

Now in order to get the direction and magnitude of the end velocity of the rock, I need both the Vx and Vy right?
Yep, that's right.

Now you have solved the first two parts. About the third part, you can go like this:-
Find the final velocity in Horizontal and Vertical direction, then find their resultant. :)

Alright, so far so good, lets see if I got this right:

To find the Velocity in the y direction near point of impact
$$V_y=V_{0y}^2+2a(y-y_0)$$

substitute in my knowns:
$$V_y=(0)-g(\frac{\sqrt{2h}}{g})$$
$$V_y^2=(0)-2g(0-h)$$
$$V_y^2=-2g(-h)$$
$$V_y^2=2gh$$
$$V_y = \sqrt{2gh}$$

As for the resultant, are you referring to vector addition?

If so, I think I got $$<V_1,0,0>+<0,\sqrt{2gh},0>\ = \ <V_1,\sqrt{2gh},0>$$

Hows all of this look?

But what if the question asked for the angle, would I need to use the Tangent Function and take the Inverse Tangent to find Theta?

Alright, so far so good, lets see if I got this right:

To find the Velocity in the y direction near point of impact
$$V_y=V_{0y}^2+2a(y-y_0)$$

substitute in my knowns:
$$V_y=(0)-g(\frac{\sqrt{2h}}{g})$$
$$V_y^2=(0)-2g(0-h)$$
$$V_y^2=-2g(-h)$$
$$V_y^2=2gh$$
$$V_y = \sqrt{2gh}$$

As for the resultant, are you referring to vector addition?

If so, I think I got $$<V_1,0,0>+<0,\sqrt{2gh},0>\ = \ <V_1,\sqrt{2gh},0>$$

Hows all of this look?

But what if the question asked for the angle, would I need to use the Tangent Function and take the Inverse Tangent to find Theta?
You got the final velocity right but with a long method.
What i expected was that you would have used the first equation of motion.

Yep, i am referring to vector addition. I think you did it by co-ordinates. I have never used that way.
We have horizontal and vertical final velocities. These both are perpendicular to each other.
So $$V=\sqrt{V_x^2+V_y^2}$$

V is the resultant velocity.
Vx=v1
Vy=$\sqrt{2gh}$

Just plug in the values and your are done.

If you want to find the angle, you need to first specify the reference, Horizontal or vertical. :)

Well, it seems I got a grasp of the problem now, I think the only thing that messed me up in the beginning was the fact that I was applying one of the formulas wrong.

Thanks for all the help, its much appreciated

Well, it seems I got a grasp of the problem now, I think the only thing that messed me up in the beginning was the fact that I was applying one of the formulas wrong.

Thanks for all the help, its much appreciated