Velocity, distance and time

In summary, the conversation involves determining the time and height of a stone thrown from a cliff and its subsequent motion. The equations used include a = v/t, Vf= Vi+at, d=Vit +(1/2)at^2, and Vf^2= Vi^2 + 2ad. Care must be taken with the direction of acceleration in calculations.
  • #1
Jenybeny27
4
0
Could someone please try and help me figure out this problem? I keep getting very large numbers and I do not think it is correct! Thanks

A stone is thrown with an initial velocity of 20 m/s straight upward from the edge of a cliff 100m above a canyon floor. The stone just misses the eliff's edge on its way down.

1.) determine the time required for the stone to reach its maximum height.
2.) determine the maximum height of the stone above the edge of the cliff.
3.) how much time elapses as the stone falls from its maximum height to the level from which it was thrown?
4.) what is the velocity of the stone upon returning to the level from which it was thrown?
5.) Determine the velocity of the stone 6 seconds after it is thrown.
6.) determine the position of the stone 6 seconds after it is thrown.

Sorry its long, but pelase help!
 
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  • #2
Show us your work so we can check it out.
 
  • #3
okay!

1.) Vi= 20m/s (given)
Vf= 0m/s (because @ max height, Vf is 0)
a = 9.81m/s^2

a= change in v/t
9.81m/s2=20m/s / t
t = 2.04 seconds (I think i did that right.)

2.) v=20m/s t=2.04s d=vt d=(20m/s)(2.04s) d= 40.8 m

((i have more work, i just want to make sure that's correct before i proceed to type it all))
 
  • #4
3.) how much time elapses as the stone falls from its maximum height to the level from which it was thrown?
a=9.81m/s2 d= 40.8m d=(1/2)at^2 40m=(1/2)9.81m/s2(t^2) t= 2.85s
 
  • #5
Jenybeny27 said:
1.) Vi= 20m/s (given)
Vf= 0m/s (because @ max height, Vf is 0)
a = 9.81m/s^2

a= change in v/t
9.81m/s2=20m/s / t
t = 2.04 seconds (I think i did that right.)
Right!

2.) v=20m/s t=2.04s d=vt d=(20m/s)(2.04s) d= 40.8 m
Incorrect. The equation d = vt assumes the speed is constant, but that's not true here. The motion is uniformly accelerated. What are some useful kinematic equations for uniformly accelerated motion?
 
  • #6
Jenybeny27 said:
3.) how much time elapses as the stone falls from its maximum height to the level from which it was thrown?
a=9.81m/s2 d= 40.8m d=(1/2)at^2 40m=(1/2)9.81m/s2(t^2) t= 2.85s
Incorrect, because you are using the incorrect distance. (Also realize that the time it takes for the stone to rise to maximum height must equal the time it takes to fall back down to that same starting point.)
 
  • #7
Doc Al said:
What are some useful kinematic equations for uniformly accelerated motion?

Um, the equations i ahve are:
a = v/t
Vf= Vi+at
d=Vit +(1/2)at^2
Vf^2= Vi^2 + 2ad

I think I would use the d= vit + (1/2)at^2 equation, right?

Therefore, d= (20m/s)(2.04s) + (1/2)(9.81m/s^2)(2.04s)^2
d= 40.8 + 20.4
d= 61.2m (maybe?)
 
  • #8
Jenybeny27 said:
Um, the equations i ahve are:
a = v/t
Vf= Vi+at
d=Vit +(1/2)at^2
Vf^2= Vi^2 + 2ad
All good.

I think I would use the d= vit + (1/2)at^2 equation, right?
Right!

Therefore, d= (20m/s)(2.04s) + (1/2)(9.81m/s^2)(2.04s)^2
d= 40.8 + 20.4
d= 61.2m (maybe?)
Careful about directions (which are represented by signs). If you take up as positive, then the initial velocity is +20 m/s (since it's upward) but the acceleration is -9.81 m/s^2 (since it acts down). Do it over with the correct sign for a.
 

1. What is velocity?

Velocity is a measure of how fast an object is moving in a particular direction. It is a vector quantity, which means it has both magnitude (speed) and direction.

2. How is velocity different from speed?

Velocity and speed are often used interchangeably, but they have a slight difference. Speed only measures how fast an object is moving, while velocity also takes into account the direction of the movement.

3. What is the relationship between distance, time, and velocity?

The relationship between distance, time, and velocity can be described using the formula: velocity = distance/time. This means that velocity is directly proportional to distance and inversely proportional to time. In other words, the greater the distance traveled in a given time, the higher the velocity.

4. How do you calculate average velocity?

To calculate average velocity, you need to divide the total distance traveled by the total time taken. For example, if an object travels a distance of 100 meters in 10 seconds, the average velocity would be 100m/10s = 10m/s.

5. Can velocity be negative?

Yes, velocity can be negative. A negative velocity indicates that the object is moving in the opposite direction of its positive velocity. For example, if an object has a velocity of -10m/s, it means it is moving in the opposite direction of an object with a velocity of 10m/s.

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