How Does Projectile Motion Determine Maximum Height and Time to Peak?

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Homework Help Overview

The discussion revolves around projectile motion, specifically focusing on determining maximum height and time to peak for a kicked ball and a projectile fired from a gun. Participants are exploring the relationships between initial velocity components and the equations of motion under the influence of gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to calculate maximum height and horizontal components of initial velocity using projectile motion equations. There are questions about the setup of formulas and the values used in calculations.

Discussion Status

There is a mix of attempts to clarify the use of formulas and check calculations among participants. Some guidance has been offered regarding the equations of motion, but there is no explicit consensus on the correct answers or methods being used.

Contextual Notes

Participants are working under the assumption of neglecting air resistance and are referencing previous questions for context. There is some confusion regarding the correct application of initial velocity components and the resulting calculations.

Musicman
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A child kicks a ball with an initial velocity of 8.5 meters per second at an angle of 35° with the horizontal, as shown. The ball has an initial vertical velocity of 4.9 meters per second and a total time of flight of 1.0 second. [Neglect air resistance.]
what is the max height? what is the horizontal componenet of the ball's initial velocity?

A projectile is fired from a gun near the surface of Earth. The initial velocity of the projectile has a vertical component of 98 meters per second and a horizontal component of 49 meters per second. How long will it take the projectile to reach the highest point in its path?

please show formula and answer.

thanks, and these are my last questions bw.
 
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You're going to use the same equations as the last question you asked.

Remember:

V(t) = At + Vo
X(t) = (1/2)At^2 + volt + Xo

Vo = Initial Velocity
Xo = Initial Position

These are good for any axis.

You should try going through the problem real quick and looking at how your past questions were solved.
 
ok for the first problem was ur max height 4.9 m.

and for problem two did u get 7 m/s as the horizontal component of the initial point? u don't have to tell me the answer if its wrong, jus tell me if its wrong because i think i put a negaive where i shouldn't have
 
I think something is wrong with your first answer. How are you setting up your formulas?

I got 7 m/s for the horizontal component though.
 
nvm i got 8.5 for the first one, but if it isn't that then it is 1.2 it depends which i was supposed to use for the firs component which I am still confused about
 
One of those answers is what I got. Show me how you're setting up your formulas.
 
my Voy=4.9, my V=8.5

so i did -5.73=-12.1+9.8t, 6.37=9.8t, =0.65 SO

then i used the 3rd equation so yea its 1.2m
 
I'm a little confused about how you're using the formulas but you came out with the right answer. I'll show you how I did it.

Our formula for height is going to be:

Y(t) = -5t^2 + 4.9t + 0

I rounded gravity (9.8) to 10 and 1/2 of that is 5. I usually deal with gravity as being a negative force so that's why the negative is out in front (A = -10 m/s^2), 4.9 just comes from our Vo (initial velocity).

Now we know total flight time is 1 second and we know that it takes the same amount of time to go up as it does to come down so the max height will occur at .5 seconds. Plug that in for t and you've got 1.2 m.
 

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