How long will it take for the baseball to reach John again if he misses it?

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khzak1
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Homework Statement


[/B]
1. Rob throws a baseball upwards at 14.2 m/s. His friend, John, is sitting in a tree 4.5m above Rob.

a. Calculate how long it will take to reach John.

b. If John misses the ball as it moves upwards, how long will it take to reach John again.

Homework Equations


Δd=V(Δt) + 1/2 (a)(Δt)^2

The Attempt at a Solution


I have found the time using the quadratic equation for part a which is 0.36s, but I am stuck on part b. I do not understand how to do it. It seems like Δd is the same in both parts. How in hell do I find Δt for part b.
 
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khzak1 said:

Homework Statement


[/B]
1. Rob throws a baseball upwards at 14.2 m/s. His friend, John, is sitting in a tree 4.5m above Rob.

a. Calculate how long it will take to reach John.

b. If John misses the ball as it moves upwards, how long will it take to reach John again.

Homework Equations


Δd=V(Δt) + 1/2 (a)(Δt)^2

The Attempt at a Solution


I have found the time using the quadratic equation for part a which is 0.36s, but I am stuck on part b. I do not understand how to do it. It seems like Δd is the same in both parts. How in hell do I find Δt for part b.
When you solve a quadratic, how many solutions do you get?
 
Well, you get 2 roots, but one is deemed useless because it is a negative.
 
khzak1 said:
Well, you get 2 roots, but one is deemed useless because it is a negative.

If you set up your equation correctly you'll get two positive roots, one being the 0.36 s you've found, and another at some later time.
Show your working and we'll see what went wrong.
 
Your quadratic equation is correct, you've made an error somewhere while solving it.
 
khzak1 said:
-14.2 ± √113.35
-9.81

Δt=0.36s
Δt=-2.53s​
I'm guessing you made a sign error. Initial velocity is positive but acceleration is negative, so your -14.2 is the wrong sign. If you cannot spot your error, please post all steps.