One dimensional motion problem

In summary: So in summary, the problem is asking to calculate the ratio of the time the athlete spends above Ymax/2 to the time it takes him to reach that height starting from the floor. This can be done using the constant acceleration equations, with the initial velocity in the second part being the final velocity in the first part. The issue with complex roots can be resolved by using a positive value for acceleration due to gravity.
  • #1
Amar.alchemy
79
0

Homework Statement


Challenge Problem(2.96) from University Physics textbook:
In the vertical jump, an athlete starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00s in the air (their "hang time"). Treat the athlete as a particle and let Ymax be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above y/2 to the time it takes him to go from the floor to that height. You may ignore air resistance.

Homework Equations


constant acceleration equations: y=Y0 + V0t - 4.9t2...1
V2y=V20y-2g(y-y0)...2

The Attempt at a Solution


Part 1(Time it takes him to go from floor to Ymax /2):
ay= -g, origin at the floor, V0y=0, y=Ymax /2, y0=0

so if i substitute these known quantities in the second equation, then for the velocity at the position Ymax /2 i am getting complex roots. Kindly inform me where i am going wrong.

Thanks,
 
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  • #2
Try using a positive value for g. Since when you ultimately calculate time, using a negative value would result in a negative value under the square root.

Edit: Strike out what I said.
 
Last edited:
  • #3
No. if v0y=0 then how can he even get into the air? You can calculate what v0y is because you know what vymax is. What is vymax?
 
  • #4
Cyosis said:
No. if v0y=0 then how can he even get into the air? You can calculate what v0y is because you know what vymax is. What is vymax?

Thanks Cyosis, I got it. since we know that Vymax is zero we can calculate the initial velocity of the part 2 which becomes the final velocity of part 1. am i right??
 
  • #5
Yep.
 

1. What is one dimensional motion?

One dimensional motion is the type of motion that occurs along a straight line. This means that an object is only moving in one direction and its position can be described by a single coordinate.

2. What are the basic equations used in one dimensional motion problems?

The basic equations used in one dimensional motion problems are the equations of motion, which include the equations for displacement, velocity, and acceleration. These are:

Displacement (Δx) = V0t + ½at2

Velocity (v) = V0 + at

Acceleration (a) = (v - V0) / t

3. How do you solve a one dimensional motion problem?

To solve a one dimensional motion problem, you need to first identify the known and unknown variables. Then, use the appropriate equation(s) to solve for the unknown variable(s). It is also helpful to draw a diagram and label the given information to visualize the problem.

4. What are some common units used in one dimensional motion problems?

The most common units used in one dimensional motion problems are meters (m) for displacement, meters per second (m/s) for velocity, and meters per second squared (m/s2) for acceleration. However, other units such as feet, kilometers, and miles can also be used depending on the context of the problem.

5. How does one dimensional motion differ from two and three dimensional motion?

In one dimensional motion, an object is only moving in one direction along a straight line. In two and three dimensional motion, an object can move in multiple directions and its position is described by multiple coordinates. This means that equations and problem-solving methods for one dimensional motion may differ from those used in two and three dimensional motion.

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