What is the initial velocity of a ball thrown straight up?

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Homework Help Overview

The discussion revolves around calculating the initial velocity of a ball thrown straight up, given the time to reach maximum height and the acceleration due to gravity. The problem is situated within the context of kinematics.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to apply kinematic equations to derive the initial velocity based on the time to reach maximum height and the acceleration due to gravity. Some express their reasoning by equating the initial velocity with the product of acceleration and time, while others restate similar calculations.

Discussion Status

Multiple participants have arrived at the same numerical result for the initial velocity, indicating a shared line of reasoning. However, the discussion does not explicitly confirm a consensus on the approach or the assumptions made in the calculations.

Contextual Notes

Participants are working with limited information, specifically the time to maximum height and the gravitational acceleration, without additional context about the ball's mass or other potential factors influencing the motion.

chemguy990
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A ball of unknown mass is thrown straight up.

*maximum height reached in 5.78 seconds
*acceleration due to gravity 9.8 m/s2

Calculate initial velocity in m/s

Would the initial velocity be 9.8 m/s2 x 5.78 s: 56.644 m/s?

Thats all the information given.
 
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chemguy990 said:
A ball of unknown mass is thrown straight up.

*maximum height reached in 5.78 seconds
*acceleration due to gravity 9.8 m/s2

Calculate initial velocity in m/s

Would the initial velocity be 9.8 m/s2 x 5.78 s: 56.644 m/s?

Thats all the information given.
that's what i got

[tex]v_y=v_{0y}+gt[/tex]

[tex]-v_{0y}=gt[/tex]
 
Not to be redundant with the other post but i got the same=P

V initial x (Acceleration)(time) = V final

At the highest point, the velocity is 0 so

V initial + (-9.8m/s^2)(5.78s) = 0

V initial = 56.644 m/s
 
Inonin said:
Not to be redundant with the other post but i got the same=P

V initial x (Acceleration)(time) = V final

At the highest point, the velocity is 0 so

V initial + (-9.8m/s^2)(5.78s) = 0

V initial = 56.644 m/s
no prob :) so that's 3 for 56.644 m/s!
 

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