Projectile Motion Graphs and Velocity Vector Calculation Homework

[bakamanju]

Homework Statement

In each case, draw a set of graphs for the vertical and horizontal components of motion. Also draw a motion map of the vertical and horizontal components of motion.

A projectile is launched with a speed of 30 m/s at an angle of 30 degrees. At a point which is ½ the way down from it’s highest point, what is the actual velocity vector (answer includes both magnitude and direction).

Homework Equations

Vox = Vo*cos(theta)
Voy = Vo*sin(theta)
a = Voy - gt
x = Xo + Vot + 1/2(at)^2
v = Vo + at
A + B = Midpoint
2

The Attempt at a Solution

I drew a vector diagram that helped me find Vox and Voy using the equations listed above. I got Vox = 25.9808 m/s and Voy = 15 m/s.

Then, I got the maximum height at 1.23718 seconds using the Voy equation and then since the half way point is essentially the midpoint of the total time it takes for the projectile to reach the ground, I took the midpoint of the final time and the maximum height time and got 1.85577 seconds.

I'm lost on how to go from here or what I'm to do... Please help if possible?

 

[collinsmark]

Hello Bakamanju,

Welcome to Physics Forums!

I drew a vector diagram that helped me find Vox and Voy using the equations listed above. I got Vox = 25.9808 m/s and Voy = 15 m/s.

So far so good.

Then, I got the maximum height at 1.23718 seconds using the Voy equation

I think 1.23719 seconds is incorrect for the time taken to reach maximum height. But anyway, see below; this problem can be solved without solving for time at all.

by the way, one of your kinematics equations (that you have not listed) will allow you to find the maximum height (in meters) directly. Using that equation will save you a lot of work. (Hint: memorize your basic kinematics equations for uniform acceleration!)

and then since the half way point is essentially the midpoint of the total time it takes for the projectile to reach the ground, I took the midpoint of the final time and the maximum height time and got 1.85577 seconds.

I believe the problem statement is asking for the midpoint in terms of distance, not time.

And the midpoint in distance (from maximum height to ground) does not happen at the midpoint in time (from maximum height to ground). That would be true if the projectile traveled in a triangle shape, but it does not. It travels in a parabola shape.

And by the way, once you calculate the halfway point in terms of distance, there is also a kinematics equation that will let you solve for velocity directly (the y-component, that is). It is possible to solve this problem without ever solving to time at all.

 

[bakamanju]

Thank you very much for the assistance. I think I understand it now!

 

[bakamanju]

I believe I found the velocity at midpoint of the distance that is traveled by the projectile.

By using (Vfy^2 - Vo^2)/ (2*-9.8 m/s^2), I was able to calculate the maximum height because Vfy would be 0 at the apex of the projectile's distance. I got 45.92 meters for the maximum height.

Do I divide by two to find the midpoint of the vertical distance and use the same equation to solve for velocity (the y component?). I'm think I'm getting my kinematic equations mixed up again... or my basic algebra.

 

[collinsmark]

By using (Vfy^2 - Vo^2)/ (2*-9.8 m/s^2), I was able to calculate the maximum height because Vfy would be 0 at the apex of the projectile's distance. I got 45.92 meters for the maximum height.

Something is not quite right with 45.92 meters. Your formula is good, but you might not have used it quite right.

I think you might have used 30 m/s for V_0y. However, that's not it. 30 m/s is the overall magnitude of the velocity, but the y-component is only part of that. You'll have to multiply that by a trig function to get the y-component of velocity.

Do I divide by two to find the midpoint of the vertical distance and use the same equation to solve for velocity (the y component?).

Yes, that right!