Solve Free Fall Object: Find Height, Velocity & Time

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Homework Help Overview

The problem involves a ball thrown vertically upward from a height of 25m with an initial speed of 10m/s, under free fall conditions. Participants are tasked with finding the maximum height, the velocity at impact, and the time taken to reach the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using kinematic equations to solve for maximum height, impact velocity, and time of flight. There is an exploration of the signs associated with velocity and acceleration due to gravity.

Discussion Status

Some participants have made attempts to calculate the maximum height and impact velocity, while others are questioning the signs used in their calculations and the interpretation of time in the context of the problem. There is an ongoing dialogue about the correct application of kinematic equations and the directionality of vectors.

Contextual Notes

Participants are navigating the implications of choosing a reference point for time and the associated signs for velocity and acceleration. There is a recognition that time can be negative in certain contexts, prompting further discussion on the conventions used in physics problems.

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1. A ball is thrown vertically upward with an initial speed of 10m/s from thetop of a building, at a height of 25m from the ground. Assume free fall conditions.
a. To what maximum height above the ground does the ball rise?
b. With what velocity does the ball strike the ground?
c. How long does the ball take to reach the ground?


2. V=v0+at
y-y0=v0*t+1/2at^2




3. For the first part (a) i used v=v0+at and plug in my knowns
v=0, v0= 10m/s, and i solved for t which came out to be t=1.0 sec. With that i can find my maximum which I plug into the equation y-25=10(1)+(1/2)(-9.8)(1^2) and got the answer 30.1m
and for part b and c, I have no clue how to solve it if anyone can help me with a simple problem that be great
thanks for the help!
 
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Part (b) is asking for a velocity. What do you know that might allow you to calculate velocity?
 
v^2=v(int)^2+(2)(delta y)...?
 
i found v^2=10^2+2(-9.8)(-25)
did the math and v=24.29m/s when it hits the ground
 
OK, now part (c) asks you for a time. What do you know that could allow you to calculate time?
 
the first equation v=v0+(-9.8)t solve for t which i have (v-v0)/(-9.8)=t
and i got -1.45s which is totally wrong since time can't be negative, I am not surewhere i have done wrong...
 
Can I see your work? What values did you plug into the equation? (other than -9.8 m/s² of course)

By the way, time actually can be negative sometimes. Whenever you do a physics problem involving time, you have to pick some event to be t=0. If something happens before that event, it will have a negative time coordinate. It's a lot like space coordinates (x, y, z); you have to pick a point to be, say, x=0, and something that is located to the left of that point will have a negative x coordinate.

In this case I think you're right, though, that the time should be positive.
 
i have plugged in
(24.29-10)/-9.8=-1.45
 
Which direction is positive, up or down?

Once you figure that out, make sure the numbers you plug in have the correct signs.
 
  • #10
well it should be (24.29+10)/9.8 but I am not getting the concept, I know that 24.29 is the velocity when it hits the ground and 10 is the initial velocity so they have to be positive but I don't know about gravity since gravity is always pull downward so it has to be negative right...? if you can explain it to me that be great
 
  • #11
bambino321 said:
gravity is always pull downward so it has to be negative right...?
OK, so downward is negative, right?

Remember that velocity is a vector: it has a magnitude and a direction. What is the direction of the 10 m/s? What is the direction of the 24.29 m/s?
 

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