Object thrown upward with known initial and final velocity

[revere21]

Homework Statement

A cliff diver runs horizontally at 4.00 m/s. He hits the water 3.00 s later. Ignore air resistance.

(a) What is the diver's speed (magnitude of the velocity vector) just before he hits the water?

Homework Equations

Change in y-component: y(t) = vyot + (1/2)g(t^2)
Change in x-component: x(t) = vxot
V(t) = Vyo - gt

The Attempt at a Solution

To find the cliff height, I simply used -0.5(9.8)(3^2). So this comes out to 44.1 meters high.
To find the change in x-component (how far he traveled), I used x = v*t, which was (4.00)(3.00), which equals 12.0 meters. So far, super easy high school-level problem.
To find his speed, I plugged in 2.998 for time 1, and 2.999 for time 2 (two separate calculations, of course), so y(2.998) = vyot - 0.5(9.8)(2.998^2). Anyways, my answer keeps ending up as 29.4 m/s, which is one of the answers on the multiple choice test review (not a graded assignment, just a past test meant for review purposes), but the correct answer is 29.7 m/s. Regardless of how small of a time interval I calculate (i.e. if I calculate the difference between Velocity at time 2.9999998 seconds and 2.9999999 seconds) I cannot figure out where this 29.7 m/s comes from.

Please advise. Thanks everyone.

 

[SammyS]

You're using the wrong equations. x(t) means x is a function of t. It doesn't mean multiply x times t.

How does v_x, the x-component of velocity change during the 3 seconds?

How does v_y, the y-component of velocity change during the 3 seconds?

 

[revere21]

You're using the wrong equations. x(t) means x is a function of t. It doesn't mean multiply x times t.

How does v_x, the x-component of velocity change during the 3 seconds?

How does v_y, the y-component of velocity change during the 3 seconds?

Yes, I understand that x(t) means x is a function of t. I didn't multiply the two together. My apologies for being unclear on that.

But to answer your question, the velocity in the x-direction remains constant at 4.00 m/s throughout. So, for every second that passes the diver moves horizontally away from the cliff by 4 additional meters.

And the velocity in the y-direction begins at 0 and increases to the tune of (0.5)(9.8)(t^2). So, between t = 0 and t = 3, the change in y was 44.1 meters in the downward direction.

Perhaps I am misunderstanding your questions, though.

 

[SammyS]

...
But to answer your question, the velocity in the x-direction remains constant at 4.00 m/s throughout. So, for every second that passes the diver moves horizontally away from the cliff by 4 additional meters.

And the velocity in the y-direction begins at 0 and increases to the tune of (0.5)(9.8)(t^2). So, between t = 0 and t = 3, the change in y was 44.1 meters in the downward direction.

Perhaps I am misunderstanding your questions, though.

The y-component of velocity, v_y, changes by 9.8m/s every second the diver is in the air. It starts at 0. After three seconds, what is v_y?

What you are finding is how much y is changing. You're not asked for that.

 

[revere21]

The y-component of velocity, v_y, changes by 9.8m/s every second the diver is in the air. It starts at 0. After three seconds, what is v_y?

What you are finding is how much y is changing. You're not asked for that.

Okay, so the acceleration of gravity is 9.8 m/s², which means that gravity causes an object to fall 9.8 meters per second, per second. Am I understanding the gist (sp?) of the point you are trying to convey?

Still, in 3 seconds the gravity would have the object moving 3*9.8 = 29.4 m/s, which is the same as the answer I was calculating (albeit in a long, unnecessarily drawn-out way). So is the answer key answer of 29.7 m/s incorrect, or am I missing something?

I appreciate your efforts in helping a dumb sophomore like myself.

 

[SammyS]

Okay, so the acceleration of gravity is 9.8 m/s², which means that gravity causes an object to fall 9.8 meters per second, per second. Am I understanding the gist (sp?) of the point you are trying to convey?

Still, in 3 seconds the gravity would have the object moving 3*9.8 = 29.4 m/s, which is the same as the answer I was calculating (albeit in a long, unnecessarily drawn-out way). So is the answer key answer of 29.7 m/s incorrect, or am I missing something?

I appreciate your efforts in helping a dumb sophomore like myself.

29.4m/s is the y component of velocity the moment before hitting the water. The x component is still 4.00 m/s.

Now, find the magnitude of the velocity. That's the speed.

 

[revere21]

29.4m/s is the y component of velocity the moment before hitting the water. The x component is still 4.00 m/s.

Now, find the magnitude of the velocity. That's the speed.

Oh my goodness, thank you. Can't believe I didn't catch that one. Very, very simple problem, as it turns out.

Thanks, SammyS. Awesome resource to have when my mind is fried from studying.