Projectile Motion: Horizontal Distance and Relative Speed Calculation

[sharp531]

Homework Statement

A gull is flying horizontally 8.00 m above the ground at 6.00 m/s. The bird is carrying a clam in its beak and plans to crack the clamshell by dropping it on some rocks below. Ignoring air resistance, a. what is the horizontal distance to the rocks at the moment that the gull should let go of the clam? b. With what speed relative to the rock sdoes the clam smash into the rocks? c. With what speed relative to the gull does the clam smash into the rocks?

Homework Equations

I. y = y0 + Voy*T - .5gt^2
II. Vy = Voy - gt

The Attempt at a Solution

I figured out part A by first finding the time from equation I(because Voy is zero and y is zero), and then taking the time and multiplying it by the relative velocity of the gull. For part B and C, I am not sure if equation II. is needed. I've tried solving it for the final speed. But if they ask for speed, which is total distance / time, would you need a velocity equation? Since this is projectile motion(free fall), I can't think of what else to do. I'm not sure if any vector calculations are necessary. Any suggestions would be great. Thanks.

 

[tiny-tim]

Welcome to PF!

Hi sharp531! Welcome to PF!

I don't think you've got part A right.

Try part C first … it's the easiest one …

show us your full calculations, and then we can see what went wrong, and we'll know how to help!

 

[sharp531]

alright, so for part a. this is what I did. First I took the formula y = yo + Voy + .5at^(2). yo is zero. The Voy is 8.00 m. Acceleration is -9.8 m/s^(2) and time is what I solved for. Time equals 0.816 seconds. Then I took the time and multiplied it by 6.00 m/s to get 4.89 m which is the horizontal distance to the rocks at the moment that the gull should let go of the clam. I have tried some things for the other two parts but can't figure it out.

 

[pgardn]

Maybe you should ask yourself these questions:

1. How long does it take to fall 8 meters from rest (Vertically-y the clam starts from rest, yes?)
I think it takes longer that 0.8 seconds...

2. What is the Horizontal x-velocity the entire time the clam is falling?

This should get you through part a. because you will have time and velocity in the horizontal.

I think part b and c may be asking you about the horizontal and vertical velocity of the clam.
So you already have the answer to one of them... (remember if you disreguard air resistance what you can say about the Horizontal x- velocity the entire time the clam is falling?) For the vertical y-velocity, use your kinematics and remember that initial velocity in the vertical y-direction is zero.

Same thing as pushing a clam off an 8 meter table with a 6 m/s initial velocity in the horizontal direction, yes? You did not give it any vertical velocity to begin with... gravity will do that for you.

 

[sharp531]

thanks for the help. This is weird because you are right that it can't be 0.8 seconds, however I was still able to use that as an input to calculate the right answer for a.(it is 4.6 m). I went back and tried to organize everything so I know what to do. As you said, y0 = zero because the clam starts from rest.
-the initial velocity in the horizontal x-direction is 6 m/s.
-the initial velocity in the vertical y-direction must be calculated right? I used the equation x = x0 + Vox*T + .5at^(2). I then solved by setting it equal to zero, and used the quadratic equation which would give me the time when the clam hit the ground. is this correct?
then using that time I find out the initial vertical y-velocity by using the equation Vy = Voy + at. Well the final velocity in the vertical direction will be zero I think. So that means Voy will be around -19.79 m/s.
-the final velocity in the horizontal direction must be zero because it lands on the ground??

But I have confused myself more, because to solve a you use the horizontal time and velocity. That would give you 12.12m, which is not close to 4.6m? :(

 

[pgardn]

thanks for the help. This is weird because you are right that it can't be 0.8 seconds, however I was still able to use that as an input to calculate the right answer for a.(it is 4.6 m). I went back and tried to organize everything so I know what to do.


As you said, y0 = zero because the clam starts from rest.
-the initial velocity in the horizontal x-direction is 6 m/s.
-the initial velocity in the vertical y-direction must be calculated right? I used the equation x = x0 + Vox*T + .5at^(2). I then solved by setting it equal to zero, and used the quadratic equation which would give me the time when the clam hit the ground. is this correct?
then using that time I find out the initial vertical y-velocity by using the equation Vy = Voy + at. Well the final velocity in the vertical direction will be zero I think. So that means Voy will be around -19.79 m/s.
-the final velocity in the horizontal direction must be zero because it lands on the ground??

But I have confused myself more, because to solve a you use the horizontal time and velocity. That would give you 12.12m, which is not close to 4.6m? :(

Yikes hold on... I think you are making this too complicated.

We actually know that the vertical distance from the bird/clam to the ground is 8m or -8 meters if we use the clam as our starting point. The initial velocity in the y-direction must be zero because the bird is flying horizontally. The bird does not shove the claim down or toss it up while in flight it simply drops it from rest. The only reason the clam has an initial velocity in the horizontal is because it is flying at 6m/s. So the only velocity the clam has to begin with is horizontal x-direction. No initial velocity in the y-direction. Since the clam has no initial velocity in the vertical y-direction simply ask yourself how long it takes to fall -8 meters. Try this first.

And then find out for sure if the answer to a) is 4.8 meters, I don't think it is unless you got some numbers reversed in the original problem... even then I don't think this is correct.

 

[sharp531]

That's what I was initially thinking in the first post, you just have to take into account that gravity(acceleration) would be -9.8 m/s^(2). But I went onto another path for some reason. Now for the distance in a. I get 7.67 m. For part b. we know that the rocks are stationary and see the gull moving. And for part c., we have to take into account of the relative horizontal velocity of 6 m/s of the gull. Would the speed be negative because the clam is heading to the ground? At first I thought b. would just be 6 m/s or 7.67 m/s but that wasn't right.

 

[pgardn]

For part b it seems like it is simply asking what is the horizontal velocity, or the x-component of the clam's velocity as it hits the rocks. Part c seems like it asking what is the vertical velocity, or the y-component of the clam's velocity as it hits the rocks. Those should be straight forward.

B) How fast is the clam moving the entire time it is in the air in the horizontal or x- direction... How fast is the seagull moving horizontally... same thing?

c) How fast will anything dropped from rest be going in the verical direction (y-direction) if dropped from 8 meters... and this of course involves acceleration, b) does not. And since it is velocity the direction is down which is usually the negative direction in these types of problems.

 

[sharp531]

Thanks I got it now. I think I messed up on my drawing of the problem so I redid it and saw what I needed to solve for. Then b.'s velocity was just equal to the square root of 5.91^(2) + (acceleration*T)^(2). Then c. would just be the vertical velocity which was used to solve for b.

 

[pgardn]

I would check that again.

The horizontal velocity... it has to be the same as the gull. The clam would remain in the same positon horizontally as the gull while accelerating down Vertically...

Or one might also say when the clam hits the ground, the gull will be directly overhead 8 meters straight up. The gull is not falling but it is moving horizontally. The clam is moving horizontally the same velocity as the gull WHILE the clam is also falling. Mabye try to see this as two separate motions. Horizontal is constant, vertical involves acceleration. No equations with acceleration necessary for any motion in the horizontal for this one, although you did need to find the time the clam was in the air with vertical motion a equation that did involve acceleration.

 

[sharp531]

First, I made a mistake in my last reply. I'm working on two of these problems with different numbers, but same setup. So I meant to say 6.0 not 5.91. Your explanation makes sense, that the horizontal velocity calculation in b. wouldn't need acceleration. I'm not sure how I ended up with the correct answer. I treated it as solving for the resultant, and you have the other two sides of the triangle.