Introductory physics: Time for a ball thrown vertically to reach maximum height

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Homework Help Overview

The discussion revolves around a physics problem concerning the time it takes for a ball thrown vertically to reach its maximum height. Participants are examining the relationship between distance, initial velocity, and acceleration due to gravity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find distance and question the relevance of height in relation to the time calculation. There is also a focus on clarifying the problem's requirements and the variables involved.

Discussion Status

The conversation is ongoing, with some participants providing calculations and others seeking clarification on the problem's specifics. There is acknowledgment of potential misunderstandings regarding the variables used and the nature of the question being asked.

Contextual Notes

Some participants note the absence of the full problem statement and express confusion regarding the values and terms used in the calculations. There is an indication that images shared may not be accessible to all participants.

danielsmith123123
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Homework Statement
A 3 kg ball is thrown vertically into the air with an initial velocity of 15 m/s. What is the time it takes for the ball to reach its maximum height?
Relevant Equations
Vf =vi +at
Vf^2 = Vi^2 +2ax
x = Vf t - (1/2)(a)(t^2)
x = Vi t + (1/2)(a)(t^2)
Is the answer key wrong? I keep getting the same answer and it is verified with the freefall equation distance=1/2 (g)(t^2)
IMG_5721[1120].PNG
 
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You should write the full text of the problem, not an abstract of it. Where did the value of d comes from?
 
Ok thank you, i edited the forum. I guess you can't open pictures on this website, but i calculated d with Vf^2=Vi^2 +(2)(a)(d)
0^2 = 15^2 + (2)(-9.8)d
d = 11.4m
(I realize i probably shouldn't use "x" and "d" interchangibly)
 
You found the maximum height. The problem is asking you to find the time it takes to reach that maximum height. Your answer should be a number in seconds, not meters.
 
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danielsmith123123 said:
Ok thank you, i edited the forum. I guess you can't open pictures on this website, but i calculated d with Vf^2=Vi^2 +(2)(a)(d)
0^2 = 15^2 + (2)(-9.8)d
d = 11.4m
(I realize i probably shouldn't use "x" and "d" interchangibly)
You don't need the height but you got it OK. What is your problem? I cannot read the image you posted. It is not clear enough.
 

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