Calculating Air Time of 23.8 ft Toss: Part B

[med9546]

A. If a person is thrown to maximum height of 23.8 ft, how long does she spend in the air during the toss? I figured out this answer. It is 2.43 seconds.

B. Is the amount of time the person is above a height of 11.9 ft more that, less than, or equal to the amount of time the person is below a height of 11.9ft?
Explain.
Well, I know the answer is not equal. I don't not know what formulas to use. I am completely lost.

C. Verify your answer to part B with some calculations.
_____ seconds above 11.9 ft.
_____ seconds below 11.9 ft.

I worked part A using meters cause that what we have been doing in class. I need help with direction. I don't know where to begin for this second part. I thought of one thing but there was no formula to complete the second part so I got lost.

 

[HallsofIvy]

It would be better to show us howyou tried to do the problems as well as the answer you got.

In problem A, the only difference between "meters" and "feet" is that, while the acceleration do to gravity is 9.81 m/s², it is 32.2 feet/s².

I presume you are using the formula h= (-g/2)t²+ V₀t+ h₀ where V₀ is the initial speed and h₀ the initial height. You may be taking h₀= 0.

In that case, the maximum height is given by V₀²/g. Knowing the maximum height you can figure out what V₀ is and then find the rest of the answers by solving the equation (-g/2)t²+ V₀t= 11.9 for t.

 

[wais]

A. Great job on solving part A! To answer part B, we need to use the equation for the height of an object in free fall: h(t) = -16t^2 + vt + h0, where h(t) is the height at time t, v is the initial velocity, and h0 is the initial height. In this case, we know that the maximum height reached is 23.8 ft, so h(t) = 23.8 ft. We also know that the initial height is 0 ft (since the person starts on the ground) and the initial velocity is 0 ft/s (since the person is thrown straight up). Plugging in these values, we get 23.8 = -16t^2 + 0t + 0, which simplifies to t^2 = 23.8/16 = 1.4875. Taking the square root of both sides, we get t = 1.22 seconds. This means that the person spends 1.22 seconds going up and 1.22 seconds coming down, for a total of 2.44 seconds in the air.

B. The amount of time the person is above 11.9 ft is equal to the amount of time the person is below 11.9 ft. This is because the person reaches a maximum height of 23.8 ft, which is exactly double the height of 11.9 ft. Since the person's motion is symmetrical, the time spent above and below 11.9 ft will also be equal.

C. To verify our answer, we can use the same equation as before. However, this time we will plug in the height of 11.9 ft and solve for t. This gives us 11.9 = -16t^2 + 0t + 0, which simplifies to t^2 = 11.9/16 = 0.74375. Taking the square root, we get t = 0.86 seconds. This means that the person spends 0.86 seconds above 11.9 ft and 0.86 seconds below 11.9 ft, which confirms that the time spent above and below 11.9 ft is equal.