Need help on freefall by 10:00 EST tonight Assignment is due today.

[Bensky]

This is due by 10:00 tonight, I have gotten all the problems right except for the last one which I cannot figure out. If anyone could help that would be nice. :) Thanks.

Homework Statement

A lit firecracker is shot straight up into the air at a speed of 54.00 m/s. How high is it above ground level 5.4000 s later when it explodes?
______ m
How fast is it moving when it blows up?
______ m/s
How far is it from its maximum possible height? Ignore drag and take g equal to 9.800 m/s
______ m

Homework Equations

highest point - v^2 = vo^2+2g(x-x0)
v=d/t

The Attempt at a Solution

v^2=vo^2 + 2g(x-x0)
0=(54^2)+2(9.8)(x)
-2916=19.6x

ignore -sign
x = 148.78 m (highest pt)

v=d/t
54=d/5.4
d= 291.6 m (for part one, doesn't make sense, greater than the highest point?)

 

[PiratePhysicist]

First find when it peaks, so with an acceleration of g=-9.8m/s^2 you'd have:
$$v = v_0 +gt$$
When v=0, then you'll have the time at the peak. If this is greater than 5.9 seconds, it's all straight forward, if it's less, then the firecracker reaches the peak, and starts falling before exploding. So you'll have to consider that.

 

[Bensky]

First find when it peaks, so with an acceleration of g=-9.8m/s^2 you'd have:
$$v = v_0 +gt$$
When v=0, then you'll have the time at the peak. If this is greater than 5.9 seconds, it's all straight forward, if it's less, then the firecracker reaches the peak, and starts falling before exploding. So you'll have to consider that.

Ok, I got ~5.51, but where are you getting 5.9 seconds from...?

 

[PiratePhysicist]

whoops, meant 5.4000 s ^_^

 

[Bensky]

OK, but I'm still confused on what you do after this. Can you tell me which formula I use to do part 1? I used one formula but it didn't accept the answer when I put it in. :\

 

[PiratePhysicist]

Well, since the time it explodes at is less than the time it peaks, we just have to plug in the explosion time $$t_e$$ into an equation that considers constant acceleration and an initial velocity. Take
$$\frac{d^2x}{dt^2}=a$$
Integrate twice plugging $$v_0$$ and $$x_0$$ in for the constants of integration and you get
$$x = x_0 +v_0 t + \frac{1}{2} g t^2$$
if plug in $$t_e$$ in for t, 54.00 in for $$v_0$$ and 0 in for $$x_0$$ you should get the height of the firecracker when it explodes.

 

[Bensky]

Well, since the time it explodes at is less than the time it peaks, we just have to plug in the explosion time $$t_e$$ into an equation that considers constant acceleration and an initial velocity. Take
$$\frac{d^2x}{dt^2}=a$$
Integrate twice plugging $$v_0$$ and $$x_0$$ in for the constants of integration and you get
$$x = x_0 +v_0 t + \frac{1}{2} g t^2$$
if plug in $$t_e$$ in for t, 54.00 in for $$v_0$$ and 0 in for $$x_0$$ you should get the height of the firecracker when it explodes.

I did this:
$$x = x_0 +v_0 t + \frac{1}{2} g t^2$$
x = 0 + (54)(5.4)+(1/2)(9.8)(5.4^2)
x= 434.484 m

rounded that to 434.5 m and the computer told me it was wrong. :\ Did I do anything wrong?

 

[PiratePhysicist]

You have the remember the sign on g, your equation has the firecracker speeding up as time goes on, it should be slowing down, so set g=-9.80

 

[Bensky]

You have the remember the sign on g, your equation has the firecracker speeding up as time goes on, it should be slowing down, so set g=-9.80

Wow, that was dumb of me. I sat here for like 30 minutes trying to figure out why this wasn't working haha.

Thanks for the help! I can probably figure out the rest easily, i'll respond here if I have any more questions..