Finding Maximum Height in a Projectile Trajectory

[hakojackie]

The question states; A projectile is fired at an angle of 60 degrees above the horizontal with an initial speed of 30.0m/s. How long does it take the projectile to reach the highest point in its trajectory? Ok so I was able to find the maximum point (34.44m) by finding the Y component. (25.98m/s)but when I plug my numbers in the equation I do not come out with the correct answer. The equation I am trying to use is Y=Voy(t)+(1/2)(-9.8m/s^2)(t^2). can anyone help me find my problem?

 

[Hootenanny]

Your maximum point is correct. Perhaps if you show your working, the formula you quoted it valid btw.

 

[tony873004]

You don't need that long formula. All you need to know is the y-component of the initial velocity and the rate of acceleration (9.8).

How long will it take something that is traveling at 25.98 m/s to be slowed to a stop if it is losing 9.8 m/s of velocity?

 

[hakojackie]

To answer Tony I understand what you are saying but I have not learned an equation or know of one to figure that out.
To answer Hootenanny this is what I get when I plug my numbers in: 34.44m=(25.98m/s)t+(-4.9m/s^2)t^2. then I solve for t:0=-34.44m+t(25.98m/s-4.9m/s^2(t)) next I get 0=-34.44m+25.98m/s-4.9m/s^2(t) then I get stuck.

 

[Saketh]

To answer Tony I understand what you are saying but I have not learned an equation or know of one to figure that out.
To answer Hootenanny this is what I get when I plug my numbers in: 34.44m=(25.98m/s)t+(-4.9m/s^2)t^2.

Looks good so far. I'm not checking your arithmetic, though.

then I solve for t:0=-34.44m+t(25.98m/s-4.9m/s^2(t)) next I get 0=-34.44m+25.98m/s-4.9m/s^2(t) then I get stuck.

How do you jump from the first axiom to the conclusion? This is an algebra error. You can just use the quadratic formula for the first equation that you got by plugging numbers in.

 

[tony873004]

I tried to phrase it so that you would intuitively come up with the equation.

For example, How long will it take something that is traveling at 10 meters per second to come to a rest if for each second that elapses, its speed reduced by 10 meters per second? That's easy. 1 second.

How long will it take something that is traveling at 20 meters per second to come to a rest if for each second that elapses, its speed is reduced by 10 meters per second? That's easy. 2 seconds.

How long will it take something that is traveling at 30 meters per second to come to rest if for each second that elapses, its speed is reduced by 10 meters per second? Any idea?

How did I arrive at 1 second using the numbers 10 and 10? How did I arrive at 2 seconds using the numbers 20 and 10? Can you write the formula I used to do the above 2 examples? Your formula will start out as " t= ".

The long equation will give you the same answer, but its more work. To do it with the long equation, before you plug in any numbers, re-write the equation to solve for t. You're plugging in your numbers first so you end up with stuff like "34.44 = 25.98...". If you re-write to solve for t, you end up with "t=..." and that's much easier to deal with.

Since you've got a t and a t², you'll need the quadratic equation, unless you take advantage of the fact that the object will take as long to drop as it took to rise. Then you can consider the final velocity to be 25.98 and the initial velocity to be 0. And since anything times 0 = 0, you can remove v₀t from your equation. Then you're just left with t², and no quadratic formula necessary.

 

[hakojackie]

Ok so I used the Quadriatic equation and got -1.1 s which is wrong due to the fact that time cannot be negative and I got 6.4s and that is not one of my choices the closest answer to choose from is 6.2s.

 

[hakojackie]

the equation you are using is initial velocity divided by final velocity. When I do this I get 2.7s which is an answer, but now my question is why don't I get that using the other equation? Shouldn't they be the same?

 

[hakojackie]

Ok nevermind I rearranged the equation like you said and got the right answer.

 

[tony873004]

...the equation you are using is initial velocity divided by final velocity. When I do this I get 2.7s which is an answer...

I imagine you meant initial velocity divided by acceleration, since that's 2.7. Initial velocity / final velocity makes you divide by 0.