I on finding seconds from maximum height.

AI Thread Summary
The discussion focuses on calculating the time it takes for a human cannonball, fired at a 45-degree angle with a muzzle velocity of 107 km/h (converted to 29.71 m/s), to reach maximum height. The correct approach involves recognizing that at maximum height, the vertical component of velocity is zero. Participants suggest using the equation of motion for vertical velocity, emphasizing the need to separate the vertical and horizontal components. The initial attempts to calculate the time using incorrect assumptions about the angle and components led to confusion. Clarifying the equations and focusing on the vertical motion is essential for finding the accurate time to reach maximum height.
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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