Calculate the speed with which the ball hits the ground

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Homework Help Overview

The problem involves a mechanics question where a ball is thrown vertically upwards from a height of 2 meters with an initial speed of 10 m/s. Participants are tasked with calculating the time the ball remains 3 meters or more above the ground and the speed at which it hits the ground.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the appropriate use of the equation s = ut + 1/2 at² and question the correct value of s to use in their calculations. There is also mention of using the quadratic equation to find time and exploring the displacement involved in the problem.

Discussion Status

The discussion is ongoing, with participants providing guidance on interpreting the problem and clarifying the variables involved. Some have made calculations and are seeking confirmation or correction of their reasoning, particularly regarding the displacement values and the application of equations.

Contextual Notes

Participants are navigating the specifics of the problem setup, including the initial height and the definitions of displacement in the context of the ball's motion. There is a focus on ensuring the correct application of physics principles without reaching a definitive conclusion.

discombobulated
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I need some help with this mechanics question please,

A ball is thrown vertically upwards with speed 10ms/s from a point 2m above horizontal ground.
a) calculate the length of time for which the ball is 3m or more above the ground.
b) calculate the speed with which the ball hits the ground.

so..i think i have to use the equation s=ut + 1/2 at2
u= 10
a= -9.8
but I'm confused about s, should it be 2m or 3m?
 
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discombobulated, this is an Introductory Physics kind of problem, not a Precalculus Mathematics one. :)
but I'm confused about s, should it be 2m or 3m?
That's neither 2m nor 3m.
s=ut + 1/2 at2
This equation will return the displacement the object has travelled.
The u is the initial velocity of that object, t is the time the objevt has spent travelling, a is its acceleration.
Now, if the ball goes from 2m above the ground to 3m above the ground. What's its displacement? What's s?
Having s, u, a, you'll be able to find t, right?
From there, do you know how long does the ball is 3m or more above the ground?
Can you go from here?
Can you do part b?
 
Thanks, so s=1 then I worked out t = 1.82 by using the quadratic equation to solve and finding the difference.
I think i need to use v2 =u2+2as for part b, but which value of s should i use?
I tried 2m but then i got v=6.26 whichis wong as the correct answer is 11.8.
 
Last edited:
discombobulated said:
Thanks, so s=1 then I worked out t = 1.82 by using the quadratic equation to solve and finding the difference.
I think i need to use v2 =u2+2as for part b, but which value of s should i use?
I tried 2m but then i got v=6.26 whichis wong as the correct answer is 11.8.
The answer you got for a looks good. But don't forget to add the measure unit to t, it should read: t = 1.82 s, not just t = 1.82.
No s is not 2. Let's think about it. You choose a = g = -9.8 m / s2, which means the positive y is upward.
So from 2m above the ground, the ball flies up, reaches its highest point, then falls back down and finally reaches the ground. So from 2m above the ground, the ball finally reaches the ground, what's the displacement, is it positive 2m? Or what is it?
Can you go from here? :)
 

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