I on finding seconds from maximum height.

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically analyzing the time taken for a human cannonball to reach maximum height after being fired from a cannon at a specific velocity and angle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to determine the time to reach maximum height using various equations of motion and questioning the role of the angle in the calculations. Some are exploring the relationship between vertical velocity and time, while others are checking the correctness of their equations.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the assumptions behind their calculations. Some guidance has been offered regarding the use of velocity components and the significance of reaching zero vertical velocity at maximum height.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also uncertainty about the correct equations to use for this specific scenario.

Sneakatone
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A human cannonball was fired from a cannon with a muzzle velocity of 107 km/h. The firing angle was 45 degrees from the horizontal


Ive converted 107 km/h to 29.71 m/s
1.How many seconds did it take for the human cannonball to reach maximum height?

2.How high did he rise?
h = (V²/4g)=22.5m
3.How far from the cannon did he land?
R = (V²/g) =29.72^2/9.8=90.1m


I cannot seem to find the time for #1 , i tried v/g=29.72/9.8=3.029 and that's wrong,
t = (V/g)√2= 4.28 and that is wrong.
what do I need to do
 
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Sneakatone said:
A human cannonball was fired from a cannon with a muzzle velocity of 107 km/h. The firing angle was 45 degrees from the horizontal


Ive converted 107 km/h to 29.71 m/s
1.How many seconds did it take for the human cannonball to reach maximum height?
You'd be seeking the time that passes until the vertical component of his velocity equals 0.
2.How high did he rise?
h = (V²/4g)=22.5m
What equation of motion did you apply here?
 
I forgot but I got 2 and 3 correct, Is there a specific equation for 1?
 
Think about what max height means. As stated above, at the maximum height, the x-component of the velocity is ZERO. Try using an equation of velocity with respect to time and keep in mind that at the maximum height, the velocity is equal to 0. Plug that into the equation along with all other known variables and solve.
 
would the angle be a factor when finding time?
I used v = gt + vi which ended up to v/g=t -> 29.72/9.81=3.029s but it is incorrect
 
I'm not completely sure, but I assume if you use the equation Vx=V0x+axt and break V0x into the component Vx=V0cos\theta you can make the equation:
V_x=Vocos(/theta)+a_xt, you can make V_x zero, plug in the rest, then solve for t.
 
Sneakatone said:
would the angle be a factor when finding time?
I used v = gt + vi which ended up to v/g=t -> 29.72/9.81=3.029s but it is incorrect
The equation is correct, but it applies to only the vertical component of velocity.
 

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