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

In summary, Rob threw a baseball upwards at 14.2 m/s and it will take 0.36s to reach his friend, John. If John misses the ball, it will take 2.53s to reach him again.
  • #1
khzak1
7
0

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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  • #2
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?
 
  • #3
Well, you get 2 roots, but one is deemed useless because it is a negative.
 
  • #4
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.
 
  • #5
-14.2 ± √113.35
-9.81

Δt=0.36s
Δt=-2.53s​
 
  • #6
Your quadratic equation is correct, you've made an error somewhere while solving it.
 
  • #7
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.
 

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

1. What is constant acceleration in 1-D?

Constant acceleration in 1-D refers to the motion of an object in one dimension (typically along a straight line), where the object's velocity changes by the same amount every second. This means that the object's acceleration remains constant throughout its motion.

2. How is constant acceleration in 1-D calculated?

To calculate constant acceleration in 1-D, the equation a = (vf - vi)/t is used, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken. This equation assumes that the acceleration remains constant throughout the motion.

3. What is the difference between constant acceleration and uniform motion?

Constant acceleration and uniform motion are two different types of motion. Uniform motion is when an object moves at a constant speed in a straight line, while constant acceleration involves a change in the object's velocity over time. In other words, constant acceleration is a type of motion where the object's speed is changing at a constant rate.

4. What are some real-life examples of constant acceleration in 1-D?

Some real-life examples of constant acceleration in 1-D include objects falling due to gravity, a car accelerating from a stop, and a rocket taking off. In all of these cases, the acceleration remains constant throughout the motion of the object.

5. How does constant acceleration affect an object's displacement and velocity?

Constant acceleration affects an object's displacement and velocity by increasing or decreasing them at a constant rate. This means that the object's velocity will change by the same amount every second, resulting in a linear increase or decrease in displacement over time.

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