# Vertical Kinematics: Solving for Jump Time

• sprinter08
In summary, the kangaroo jumped to a vertical height of 2.8m, and it was in the air for 1.14 seconds.
sprinter08

## Homework Statement

A kangaroo jumps to a vertical height of 2.8m. How long was it in the air before returning to Earth?

## Homework Equations

delta y= Vot + 1/2at^2

## The Attempt at a Solution

I had to get t by itself so I rearranged the equation to say: t= delta y/.5(9.8) all under a square root. First off, I'm not sure if I set up the equation correctly, and secondly, I am not sure if just the top (delta y) goes under the square root or both the top and bottom(or the whole side of the equation) goes under the square root.

Not sure of your approach here. What's Vo?

Try this: Start at the top. Figure out how long it takes for the kangaroo to fall a distance of 2.8m. What would Vo be in that case? How would that time relate to the total time the kangaroo was in the air?

Vo in this case would be zero, I believe...and if it would be zero, then it would eliminate the first part of the equation (Vot).

Oh, sorry, I misread your question I think. I'm not sure if I know how to figure that out.

sprinter08 said:
I had to get t by itself so I rearranged the equation to say: t= delta y/.5(9.8) all under a square root.
First off, I'm not sure if I set up the equation correctly, and secondly, I am not sure if just the top (delta y) goes under the square root or both the top and bottom(or the whole side of the equation) goes under the square root.

Looks ok to me

Vo = 0, and y = (1/2)at^2, so t^2 = 2y/a, so t = ± √(2y/a).

You seem to have the technique right - why are you unsure?

well, I was just wondering if I were to take the square of the top and then divide it by the bottom, or if I divided the top by the bottom and then took the square root. I divided the top by the bottom, and then took the square root to get .57. But then I think I would need to multiply it by 2 to get 1.14 seconds. I'm not exactly sure.

sprinter08 said:
Vo in this case would be zero, I believe...and if it would be zero, then it would eliminate the first part of the equation (Vot).
That's right.

sprinter08 said:
Oh, sorry, I misread your question I think. I'm not sure if I know how to figure that out.
Sure you do. That's exactly what you're doing! (I don't see any Vot in your final equation.)

sprinter08 said:
well, I was just wondering if I were to take the square of the top and then divide it by the bottom, or if I divided the top by the bottom and then took the square root. I divided the top by the bottom, and then took the square root to get .57. But then I think I would need to multiply it by 2 to get 1.14 seconds. I'm not exactly sure.
You divide first, then take the square root--just like you did. And the reason you need to multiply by two, is that you only solved for the time it takes to fall from the highest point (or reach the highest point). That's only half of the total time.

Thank you!

## 1. What is vertical kinematics?

Vertical kinematics is the study of objects moving in a vertical direction, such as jumping or projectile motion.

## 2. How do you calculate jump time?

Jump time can be calculated using the equation t = √(2h/g), where t is the time in seconds, h is the height in meters, and g is the acceleration due to gravity (9.8 m/s²).

## 3. Can you solve for jump time without knowing the height?

Yes, you can solve for jump time without knowing the height by using the equation v = gt, where v is the final velocity and t is the time. This only works if the initial velocity is zero and the object is only moving in the vertical direction.

## 4. What are the units for jump time?

Jump time is typically measured in seconds, but it can also be expressed in milliseconds or other units of time depending on the situation.

## 5. How does air resistance affect jump time?

Air resistance can affect jump time by slowing down the object's vertical velocity, thus increasing the time it takes for the object to reach its peak height. However, for most everyday jumps, air resistance is negligible and can be ignored in calculations.

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