What is the time when the ball is 15 m above the ground?

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Homework Help Overview

The problem involves a ball thrown upward with an initial velocity, seeking to determine the time it reaches a height of 15 meters above the ground. The subject area is kinematics, specifically projectile motion.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the kinematic equation for height but expresses uncertainty about incorporating the maximum height into their calculations. Some participants suggest setting the height equation equal to 15 meters and solving for time.

Discussion Status

Participants are exploring different methods to approach the problem, with some guidance provided on using the height equation. There is no explicit consensus on the best method yet, as the original poster is still grappling with the setup.

Contextual Notes

The original poster has indicated they know the answer but are struggling with the calculations, suggesting a potential gap in understanding the application of the kinematic equations.

carltonblues
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1)A ball is thrown upward with an initial velocity of v = 20 m/s. How long is the ball in the air? What is the greatest height of the ball? When is the ball 15 m
above the ground?


1) I have figured out

t total = 4.08s
greatest height = 20.4m

For the last bit (Finding the time when x=15m), I am having trouble. I know the answer but cannot work it through.

Do I use s = ut+1/2at^2 for it?

If I do, it won't work out because the 20.4 m has to come into it, doesn't it?

Please help!
 
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The equation giving height (let's label it through 'y') as function of time is
[tex]y(t)=v_{0}t-\frac{1}{2}gt^{2}[/tex]

Now set "y" to 15m and solve for "t"...

Daniel.
 
dextercioby said:
The equation giving height (let's label it through 'y') as function of time is
[tex]y(t)=v_{0}t-\frac{1}{2}gt^{2}[/tex]

Now set "y" to 15m and solve for "t"...

Daniel.
Cheers mate
 

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