Physics Kinematics Questions

In summary, the conversation discusses two problems related to AP Physics B homework. The first problem involves finding the takeoff speed of a professional NBA athlete with a reported vertical leap of 47.3 inches. The second problem involves calculating the time it takes for a camera dropped from a hot air balloon descending at a rate of 2.6 m/s to reach the ground from a height of 45 m. The conversation also discusses the equations and concepts necessary to solve these problems, such as the equation vf^2-vi^2=2ad for the first problem and the concept of objects in freefall for the second problem.
  • #1
ama
15
0
I have 2 problems that I am completely stuck on. I have tried to solve them, but just can't. I am working on summer homework for AP Physics B and I am stuck on these 2.

Homework Statement


The vertical leap of a professional NBA athlete is reported to be 47.3 inches. What is his takeoff speed?

displacement=47.3 inches since it is the leap

Homework Equations


can't find any


The Attempt at a Solution


I do not know what to do. I assume that the takeoff speed is the initial velocity, but I do not know any time values. These time values are needed for any equation I could use to solve this.

Second problem

Homework Statement


A hot air balloon is descending at a rate of 2.6 m/s when a passenger drops a camera.
If the camera is 45 m above the ground when it is dropped, how long does it take for the camera to reach the ground?
displacement= 45m since the camera has to be dropped

Homework Equations


displacement= initial velocity(time)+(1/2)(a)(time^(2))

The Attempt at a Solution


I assume initial velocity is 2.6. And the time is what has to be solved. I can't figure out what acceleration is though.That is what I need to know.
 
Physics news on Phys.org
  • #2
I do not know what to do. I assume that the takeoff speed is the initial velocity, but I do not know any time values. These time values are needed for any equation I could use to solve this.

Have you seen encountered the equation vf^2-vi^2=2ad? If not, you can derive it from vf=vi+at and d= v0*t+(1/2)(a)(t^2). Use the first equation to solve for t, substitute the result into the second, and you should get it after some algebra.

I assume initial velocity is 2.6. And the time is what has to be solved. I can't figure out what acceleration is though.That is what I need to know.

All objects in freefall accelerate at "g". (Incidentally, this probably comes close to being the first physics law ever discovered. It predates Newton by several decades.)
 
  • #3
For #1 then, what is the acceleration. And what would vf be.
 
  • #4
ama said:
For #1 then, what is the acceleration. And what would vf be.

What did ideasrule say about objects in freefall?

What do you suppose his velocity is when he's at his highest distance from the ground?
 
  • #5


As a scientist, it is important to approach problems in a systematic and logical manner. For the first problem, you are correct in assuming that the takeoff speed is the initial velocity. However, you also need to know the acceleration of the athlete in order to solve for the takeoff speed. This can be found using the kinematic equation v^2 = u^2 + 2as, where v is the final velocity (which in this case is 0 since the athlete reaches the peak of their jump), u is the initial velocity (takeoff speed), a is the acceleration, and s is the displacement (47.3 inches). Rearranging the equation to solve for u, we get u = √(v^2 - 2as). Since v = 0, we are left with u = √(-2as). The acceleration in this case is due to gravity, which is approximately 9.8 m/s^2. So, we can convert the displacement of 47.3 inches to meters (1.2 meters) and plug in the values to get u = √(-2*9.8*1.2) = 4.07 m/s. Therefore, the takeoff speed of the NBA athlete is 4.07 m/s.

For the second problem, you are correct in using the kinematic equation displacement = initial velocity(time) + (1/2)(acceleration)(time^2). The acceleration in this case is also due to gravity, so we can use the value of 9.8 m/s^2. The initial velocity is 2.6 m/s and the displacement is 45 m. Rearranging the equation to solve for time, we get t = √(2s/a) - u/a. Plugging in the values, we get t = √(2*45/9.8) - 2.6/9.8 = 3.27 seconds. Therefore, it takes 3.27 seconds for the camera to reach the ground.

In both of these problems, it is important to identify the known and unknown variables, and use the appropriate equations to solve for the unknowns. It is also important to pay attention to units and convert them if necessary. If you are still having trouble, I would recommend seeking help from your teacher or a tutor to guide you through the problem-solving process. Keep in mind that physics is a
 

1. What is the difference between speed and velocity?

Speed is a scalar quantity that refers to the rate at which an object covers distance. Velocity is a vector quantity that refers to the rate at which an object changes its position in a specific direction.

2. How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the change in time. It is represented by the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between average and instantaneous velocity?

Average velocity refers to the total displacement over a given time period, while instantaneous velocity refers to the velocity of an object at a specific moment in time.

4. How does the slope of a position-time graph relate to velocity?

The slope of a position-time graph represents the velocity of an object. A steeper slope indicates a higher velocity, while a flatter slope indicates a lower velocity.

5. What is the difference between distance and displacement?

Distance is a scalar quantity that refers to the total length an object has covered, regardless of its direction. Displacement is a vector quantity that refers to the shortest distance and direction between an object's initial and final position.

Similar threads

  • Introductory Physics Homework Help
Replies
22
Views
311
  • Introductory Physics Homework Help
Replies
4
Views
946
  • Introductory Physics Homework Help
Replies
6
Views
409
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
677
  • Introductory Physics Homework Help
Replies
6
Views
841
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top