Object thrown upward with known initial and final velocity

In summary: Cheers.In summary, the diver's speed (magnitude of the velocity vector) just before hitting the water is 29.7 m/s, with a constant horizontal velocity of 4.00 m/s and a downward y-component velocity of 29.4 m/s due to gravity.
  • #1
revere21
10
0

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.
 
Physics news on Phys.org
  • #2
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 vx, the x-component of velocity change during the 3 seconds?

How does vy, the y-component of velocity change during the 3 seconds?
 
  • #3
SammyS said:
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 vx, the x-component of velocity change during the 3 seconds?

How does vy, 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.
 
  • #4
revere21 said:
...
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, vy, changes by 9.8m/s every second the diver is in the air. It starts at 0. After three seconds, what is vy?

What you are finding is how much y is changing. You're not asked for that.
 
  • #5
SammyS said:
The y-component of velocity, vy, changes by 9.8m/s every second the diver is in the air. It starts at 0. After three seconds, what is vy?

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.
 
  • #6
revere21 said:
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.
 
  • #7
SammyS said:
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.
 

What is the equation for calculating the maximum height reached by an object thrown upward with known initial and final velocity?

The equation for calculating the maximum height reached by an object thrown upward with known initial and final velocity is h = (v02 - v2) / (2g), where h is the maximum height, v0 is the initial velocity, v is the final velocity, and g is the acceleration due to gravity.

What is the significance of the initial and final velocity in an object thrown upward?

The initial velocity determines the object's initial speed and direction of motion, while the final velocity determines the object's speed and direction of motion at its highest point. These velocities are important in calculating the object's maximum height and its motion throughout its trajectory.

How does air resistance affect an object thrown upward with known initial and final velocity?

Air resistance, also known as drag, will slow down the object as it moves upward and speed it up as it falls back down. This will result in a lower maximum height and a shorter total flight time compared to an object thrown in a vacuum.

What happens to the object's velocity at its highest point?

At the object's highest point, its velocity becomes zero. This is because the object has reached its maximum height and is about to start falling back down due to the force of gravity.

How can the initial and final velocity be used to determine the time of flight for an object thrown upward?

The time of flight can be calculated using the equation t = (v - v0) / g, where t is the time of flight, v is the final velocity, v0 is the initial velocity, and g is the acceleration due to gravity. This equation can be derived from the kinematic equations of motion.

Similar threads

  • Introductory Physics Homework Help
2
Replies
57
Views
572
  • Introductory Physics Homework Help
Replies
5
Views
885
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
870
  • Introductory Physics Homework Help
Replies
4
Views
946
  • Introductory Physics Homework Help
Replies
11
Views
1K
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
13
Views
670
  • Introductory Physics Homework Help
Replies
16
Views
974
Back
Top