Determine Velocity at each second

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Homework Help Overview

The problem involves analyzing the motion of a ball thrown straight up with an initial velocity of 30 m/s, under the influence of gravity, which is assumed to be 10 m/s². The goal is to determine the displacement, velocity, and acceleration at each second during its flight.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the time to reach the peak and the total distance traveled, but expresses difficulty in finding velocity, displacement, and acceleration at each second interval. They question whether acceleration remains constant until the peak and how to apply formulas for the other quantities.

Discussion Status

Some participants provide hints on how to approach the problem, suggesting the use of the equation Vf = Vi + a*t to tabulate velocity at each second. There is an exploration of different sign conventions for acceleration and velocity, with some participants questioning the implications of these choices on the calculated acceleration.

Contextual Notes

Participants are discussing the implications of sign conventions for velocity and acceleration, particularly in relation to the direction of motion and the point of maximum height. There is an acknowledgment of the need to clarify assumptions regarding the behavior of acceleration during the ascent and descent phases of the ball's motion.

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Homework Statement


A ball is thrown straight up at 30m/s. Assume g is 10m/s/s Show the dispacement, velocity, and acceleration at each position.


Homework Equations


Vf² = Vi² + 2aΔx
Vf = Vi + at


The Attempt at a Solution


So first I wanted to solve for the time it takes to go up so

Vf = Vi + at
0 = 30 + (-10)t
t = -30/-10 = 3 seconds up
then just times 2 because its the same coming down
so 6 seconds in all

Now for distance up

Vf² = Vi² + 2aΔx
Δx = (Vf² - Vi²)/2a
Δx = (0 - 900)/2(-10)
Δx = (-900)/(-20)
Δx = 45m up
and again times 2
so 90 meters in all

Now for some reason I'm having trouble figuring out how to find the velocity,displacement,and acceleration at each 1 second interval.

Should acceleration stay the same until the top where its 0 then change to +10m/s/s?

And how and what formula would I use for the other two?

Thanks in advance everyone
 
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Hint: For the velocity simply tabulate Vf = Vi + a*t. On the way up, a is negative. Use results for Vf in your first equation and solve for delta X. Repeat.
 
Thanks Lawrence ...I kept thinking I had to bring in another formula. Thanks again!
 
Actually ...when solving for acceleration ...I get that on the way up its negative ...but by using something using my results from above...
Vf = Vi + at
(Vf-Vi)/t

Using say
V(0)=30m/s and V(1)=20m/s
So
(20-30)/1 = -10 which I assumed

But

Using V(4) and V(5)

(-20 - (-10))/1
That comes to -10 as well...but shouldn't acceleration be positive at this time?
 
If you choose the starting point to be the velocity at the highest point where it is zero and let downward velocities be positive, then you will get a positive acceleration.

The equation Vf = Vi + at enables one to determine a velocity at a later time under constant acceleration over that time period. If the acceleration is in the same direction as the velocity and velocity is positive, acceleration is positive. Vice versa is also true. If you use it to determine accelerations, the sign of the acceleration is dependent on the sign convention you choose for the velocities. If the magnitude of the velocity increases, the acceleration has the same sign as the velocity. If magnitude of velocity decreases, acceleration has the opposite sign.
 

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