# 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.

#### Attachments

• phy4aH3P1.gif
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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 Your welcome!! And remember the table approach. Whenever you deal with projectiles, the best way is to make the table I forgot to welcome you. Welcome to the board. Hope you enjoy your stay here. Before wandering on the board, have a look at the rules. They may help you.