# Homework Help: I need help on finding seconds from maximum height.

1. Feb 14, 2013

### Sneakatone

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 thats wrong,
t = (V/g)√2= 4.28 and that is wrong.
what do I need to do

2. Feb 14, 2013

### Staff: Mentor

You'd be seeking the time that passes until the vertical component of his velocity equals 0.
What equation of motion did you apply here?

3. Feb 14, 2013

### Sneakatone

I forgot but I got 2 and 3 correct, Is there a specific equation for 1?

4. Feb 14, 2013

### Yosty22

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.

5. Feb 14, 2013

### Sneakatone

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

6. Feb 14, 2013

### Yosty22

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.

7. Feb 14, 2013

### Staff: Mentor

The equation is correct, but it applies to only the vertical component of velocity.