Professor leaps off a helicopter, ball thrown in the air Kinematics

[Keldroc]

Hey all, I've been lurking for a while to try and see if these problems were solved already but I couldn't find anything so here I am posting!

Question #2:A ball is thrown straight upward from the ground. The ball passes a window 4.35 m above the ground and is seen to descend past that same window 3.68 s after it went by on the way up.

1. What is the total length of time the ball is in the air?

Homework Equations

<br /> v = v_0 + a t<br />
<br /> x = x_0 + v_0 t + (1/2) a t^2<br />
<br /> v^2 = v_0^2 + 2 a \Delta x<br />
and Quadratic formula

The Attempt at a Solution

I tried finding the velocity, max height and then plugging it in the quadratic formula but it is not the right answer. I'm not sure how else to do it? Any help would be greatly appreciated

 

[cepheid]

Welcome to PF,

Hey all, I've been lurking for a while to try and see if these problems were solved already but I couldn't find anything so here I am posting!

Homework Statement

Question #1:A physics professor, equipped with a rocket backpack, steps out of a hovering helicopter 808 m with zero initial velocity. For 9 s, she falls freely. At that time, she fires her rockets and slows her rate of descent (decelerates) at 15 m/s^2, until her rate of descent reaches 5.00 m/s. At that time, she adjusts her rockets so that her acceleration is zero and she descends the remaining distance to the ground at 5.00 m/s. Neglect air resistance. [Hint: This is a one-dimensional problem.]

1. What is the professor's altitude at the end of her free fall? -410.7 m

2. What is the professor's velocity at the end of her free fall? -88.3 m/s

3. Once the professor fires her rockets, how long does it take her to decelerate to 5.00 m/s? -5.552 secs

4. What is the total time for the professor's trip from the helicopter to the ground?

---I figured out the first 3 parts but the last part is what I'm having trouble with.

Why are the values in red negative?

For the last part, the trip can be divided into three stages:

- free fall
- deceleration
- constant velocity

All you need to do is calculate the duration of each stage. For the final stage, the amount of time it takes her to descend at a constant velocity just depends upon the her height above the ground at the beginning of that stage.

 

[Keldroc]

Ok got it, thanks! But can someone help me out with the second problem too?

 

[cepheid]

Ok got it, thanks! But can someone help me out with the second problem too?

How about this? Call the time at which the ball reaches a height of 4.35 m time t₁. Substituting this into the middle equation in your relevant equations, you only have one other unknown, which is v₀. Therefore, you can solve for v₀ in terms of t₁.

Now, you know from the inherent symmetry of the situation that the time it takes to go up from 4.35 m to the max height should be equal to the time that it takes to come back down to 4.35 m from the max height -- it's (3.68/2) = 1.84 seconds in either direction. Therefore, you can substitute your expression for v₀ in terms of t₁ into the first equation in your list of relevant equations. Here, you set v = 0 and t = t₁ + 1.84 s. This equation then gives you t₁.

 

[Keldroc]

Wow thank you very much! I finally figured them both out! Thank you once again!