Solve Simple 2D Ballistics Equations: Max Height, Velocity & Distance

• Torquescrew
In summary, the conversation discusses preparing for finals and solving a physics problem involving a man throwing a rock from the roof of a building. The problem is solved using various equations and trigonometry, and the final answer is verified to be correct. The conversation also touches on the frustration of encountering unrealistic problems in school.
Torquescrew
Well, finals are coming around. I'm reviewing stuff I thought I learned at the beginning of the semester only to find out that I don't remember what I'm doing. It's pretty easy until I forget everything.
I think I have a handle on this, but I just want some verification to make sure I'm doing it right.

Homework Statement

A man stands on the roof of a building that is 30.0 m tall and throws a rock with a velocity of magnitude 60.0 m/s at an angle of 33.0° above the horizontal. Calculate a) the maximum height above the roof reached by the rock; b) the magnitude of the velocity of the rock just before it strikes the ground; c) the horizontal distance from the base of the building to the point where the rock strikes the ground.

Homework Equations

I'm a big fan of using these:
Vf=Vo+a(t)
(delta)x=Vo(t)+.5(a)(t^2)
(delta)x=[Vf^2 - Vo^2]/2a

and some basic trig

The Attempt at a Solution

For (a), I chopped up my initial vector into x and y components.
Vox = 50.3202 m/s
Voy = 32.6783 m/s

Then, I figured I'd not only see how far up it would go, but how long it would take to get there.
For time, I used:
0=32.6783{m/s}-9.81{m/s^s}*t
I wound up with 3.3113 seconds

Next, I got the same result of 54.4278 meters (above the roof) whether I used this equation
(delta)x=Vo(t)+.5(a)(t^2) <--using the time I had previously solved for
or this one
(delta)x=[Vf^2 - Vo^2]/2a <--ignoring time altogether

Then, for (b) I added the 54.4 meters to my initial 30 and solved for a new time (this one for it's downward movement) with this one again:
(delta)x=Vo(t)+.5(a)(t^2)
84.4...=0+(.5*9.81*t^2)
Netted myself another 4.14881 seconds

Using this new time, I solved for my final y velocity with
Vf=Vo+a(t)
Vf=0+9.81*4.14881
I wound up with 40.6998 m/s
I verified by using this formula:
Vf^2=Vo^2 + 2a(delta x)
or rather
Vf^2=0+(2*9.81*84.4...)

Then, I used the handy formula for building one vector out of 2:
[(x^2)+(y^2)]^(1/2) <-- somebody is gonna' have to teach me how to type a "square root" symbol
[(50.3202^2 m/s) + (40.6998^2 m/s)]^(1/2)
I now have a vector magnitude of about 64.7 m/s

Lastly, for (c), I added my two times together
4.14881+3.3113
Very nearly 7.5 seconds in flight.

I multiplied my initial x vector by the flight time to see how far it flew before it landed.
50.3202 m/s * 7.47993 s

So it looks like this guy threw a rock 376.4 meters.

That's over three and a half football fields.
Either this guy has one heck of an arm (60 meters a second initial velocity? Madness!), or I royally jacked this up.

Am I crazy? More importantly, are my calculations correct?
Something just seems out of place and I can't put my finger on it.

Your answer is correct.

Yup, the man does have one heck of an arm.

I am a senior high school student and I hate it when there are problems that do not even try to be semi-realistic. It leads me to believe that my answer is incorrect somehow.

Heck the man in this problem broke the record pitching speed for a baseball.

1. What are 2D ballistics equations?

2D ballistics equations are mathematical formulas that describe the motion of an object in two dimensions under the influence of gravity. They take into account the initial velocity, angle of launch, and acceleration due to gravity to calculate the object's trajectory.

2. How do you solve simple 2D ballistics equations?

To solve simple 2D ballistics equations, you need to first identify the known values such as the initial velocity, angle of launch, and acceleration due to gravity. Then, plug these values into the appropriate equations and solve for the desired quantity, such as maximum height, velocity, or distance.

3. What is the max height in 2D ballistics equations?

The maximum height in 2D ballistics equations is the highest point along the object's trajectory. It is calculated by finding the vertex of the parabolic path using the initial velocity, angle of launch, and acceleration due to gravity.

4. How does the angle of launch affect the trajectory in 2D ballistics equations?

The angle of launch has a significant impact on the trajectory in 2D ballistics equations. A higher launch angle will result in a longer flight time and a longer horizontal distance traveled, but a lower maximum height. A lower launch angle will result in a shorter flight time and a shorter horizontal distance traveled, but a higher maximum height.

5. Can 2D ballistics equations be used for all objects?

No, 2D ballistics equations are only applicable to objects that are launched with a known initial velocity, angle of launch, and under the influence of constant acceleration due to gravity. They do not account for air resistance or other external forces that may affect the object's motion.

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