Projectile Motion

[MDwannabe7]

Hello, this is actually my third time taking the MCAT and I'm trying to master concepts that I never understood the first two times around. I have trouble with projectile motion - trying to figure out total time in flight, maximum height, etcetera, when I don't get exactly the variables that I need. I'm pretty bad at figuring out how to rearrange some of the equations to end up with the right information.

If I'm given the mass of an object - 100kg, the vertical component of the object's velocity as a function of time between where the object is projected and it's maximum height - Vy=17m/s, t=1.7s, angle is 45 degrees and the object is projected from a height of 10 meters. I need to figure out how much time it will take this object to reach the ground.

The object is also projected by a constant force of 10,000N applied over 0.24s. Which, I believe, translates out to an initial velocity of 24m/s...Help?

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[HawkeyePostOp]

Here's one way, all you need are Vy = 17m/s and H_o = 10m.

Find max height:

.5mv^2 = mgh

cancel m
.5v^2 = gh

h = (.5v^2)/g

find the time it takes to get to that height

h = 1/2at^2 + 17t

save this as t_1

now, to account for the 10 meters, take the height from the first calculation h, and add 10 meters to it, call this h_2

to get how much time it takes to fall in the second half, use freefall equation t_2 = (2g/h_2)^(1/2)

add t_1 to t_2

 

[drbeck]

i'm having trouble with this problem too! i'm trying to use the other equations y= Vyt + 1/2 at^2 to find the max height...is t=1.7s refering to the time it takes to get to max height? i also tried using to Vy^2= Vnoty^2 + at where Vy equals 0 in order to find the time it takes to get to max height. i'm getting different answers. shouldn't i get t=1.7s as stated?

also if i use 1/2mv^2= mgh, i get a max height of about 15m, but if i use
y= Vnotyt +1/2at^2 i get something like 44m when using t=1.7s. where am i going wrong?

 

[engineeredout]

i'm having trouble with this problem too! i'm trying to use the other equations y= Vyt + 1/2 at^2 to find the max height...is t=1.7s refering to the time it takes to get to max height? i also tried using to Vy^2= Vnoty^2 + at where Vy equals 0 in order to find the time it takes to get to max height. i'm getting different answers. shouldn't i get t=1.7s as stated?

also if i use 1/2mv^2= mgh, i get a max height of about 15m, but if i use
y= Vnotyt +1/2at^2 i get something like 44m when using t=1.7s. where am i going wrong?

D = V0t + 1/2at^2. Take the point of the object at its highest and t = 1.7 for the time to hit the ground. D = 1/2*10*1.7^2 ~ 15m. Looks like you're using v0 = 17 when its actually 0 m/s

1/2*m*v^2 = mgh == .5*100*17^2 = 100*10*h, h ~ 15m

 

[HawkeyePostOp]

MD, can you post the actual question and the answer?

 

[drbeck]

i was using Vo=17m/s because that's the original velocity to get to max height. at max height, V=0. so if t=1.7s to hit the ground FROM max height, then max height is actually 5m from the 10m starting point. is this correct? then i need to find out how long it takes to get to that 5m height? it wouldn't simply be half of 1.7s? i'm obviously not a physics pro...

 

[engineeredout]

I'm just going by what the OP said, "If I'm given the mass of an object - 100kg, the vertical component of the object's velocity as a function of time between where the object is projected and it's maximum height - Vy=17m/s, t=1.7s"

It would be helpful if he posted the original actual question instead of just paraphrased pieces of it.

 

[drbeck]

maybe we just don't understand the question b/c i thought 1.7s was to get to max height

 

[drbeck]

i'm having trouble with this problem too! i'm trying to use the other equations y= Vyt + 1/2 at^2 to find the max height...is t=1.7s refering to the time it takes to get to max height? i also tried using to Vy^2= Vnoty^2 + at where Vy equals 0 in order to find the time it takes to get to max height. i'm getting different answers. shouldn't i get t=1.7s as stated?

also if i use 1/2mv^2= mgh, i get a max height of about 15m, but if i use
y= Vnotyt +1/2at^2 i get something like 44m when using t=1.7s. where am i going wrong?

i just realized Vy^2 = Voy^2 + at is not an equation. it's actually
Vy=Voy +at...wow how's that for a mess up. i hope i don't do that on the mcat.

thus, 17m/s= at (b/c Voy=0) and t=1.7s to get to max height.

 

[tco]

It would be a lot easier if you just posted the actual problem.

If we only want to know how long it takes the object to hit the ground after being launched from a platform that is 10 meters high, I would do the following calculations. It's probably not the most efficient method, but it works.

We only care about the vertical components, since that's the only thing that effects the time of flight. All "a" values are rounded to 10 m/s.

So, using the equation vf=vi + at, and assuming that the velocity of the projectile in the vertical direction is 17 m/s after 1.7 seconds, we get 34 m/s for the initial velocity.

1/2mv^2=mgh at the apex of flight. After plug and chug, we get 57.8 m.

x=1/2at^2, so, at x=57.8 m and a=10 m/s, I'm going to estimate t as 3 seconds. We're low here, remember that. Sorry, I have no calculator. That completes the first part of the problem.

The second part is just the freefall. x=1/2at^2. So, x=67.8 m (because of the 10 m platform). I'm going to estimate t as 4 because we were low with the first estimate.

3+4=7 seconds

If any steps need clarified, hollachaboy.

 

[MDwannabe7]

Sorry Everyone, I didn't realize there was so much activity on this question, or I would have gotten back to you sooner. The question comes from a practice booklet by Kaplan and has some figures, so I will describe it as best I can. Sorry for any confusion...

Okay, the question is in regards to a circus doing a new act where they shoot a clown out of a cannon with an initial velocity at a trajectory that makes an angle of 45 degrees with the ground. The cannon accelerates the clown by applying a constant force of 10,000N over a very short time of 0.24s. The height above the ground at which the clown begins his trajectory is 10m. There is a large hoop suspended from the ceiling at just the right place for the clown to dive through as he reaches his maximum height. Gravity is 10 m/s^2 and we are shown a figure that is a graph of the vertical component of the clown's velocity as a function of time between the cannon and the hoop based on different masses of clowns.

From this figure, approximately how much time will it take for a clown with a mass of 100kg to reach the safety net located 10m below the height of the canon?

The graph indicated that the vertical component of a 100kg clown's velocity will be 17m/s and the time between between the canon and the hoop is 1.7s.


That is the whole question with all the information supplied. Hope this helps...


The answer is: 3.9s

 

[drbeck]

so this question is pretty easy if you use 1/2mv^2= mgh (kinetic goes to potential energy)...

you then find the height from 10m to max height to be 15m, add this to 10m and you get total height if 25m. then using y= V0t + 1/2at^2 with V0t at the top=0, you get 25= 1/2(10)t^2 and the time from the top to bottom is 2.2s

1.7s to get to the top + 2.2 seconds from top to bottom = 3.9s

HOWEVER, why doesn't the equation y= Vot + 1/2at^2 work to find max height? Vo should be 17m/s and t=1.7s, i only get 15m to the top if i disregard "Vot", but why disregard it?

 

[engineeredout]

HOWEVER, why doesn't the equation y= Vot + 1/2at^2 work to find max height? Vo should be 17m/s and t=1.7s, i only get 15m to the top if i disregard "Vot", but why disregard it?

Try this: Don't disregard it, but remember that acceleration is negative.

17*1.7 + .5*-10*1.7^2 ~ 15m