How Do You Calculate Projectile Velocity at Different Times?

[mgerman63016]

I would like to better understand how to work Velocity of a Projectile object. I have homework due tonight but I'm going to turn it in late to have a better understanding.

Question: A projectile is launched with an initial velocity of 39.3 m/s at 56.2° above the horizontal. Calculate the magnitude and direction of its velocity at each of the following times.

a.) 2.05 seconds after launch.
b.) 5.10 seconds after launch.
c.) Calculate the magnitude of it's velocity at 5.10 seconds
d.) What is the direction of the projectile's velocity vector 5.10 seconds after launch?

I've tried to google videos and tried to get help from the internet and nothing was coming out right. I still got everything wrong.

some guidance would be greatly appreciated PLEASE.

 

[tms]

The first step is to write down the relevant equations.

 

[FeynmanIsCool]

Khan Academy videos on youtube- great!

 

[mgerman63016]

relevant equations like:

the vertical (y) picture for this setup
which I came up with

Y =
V_oy = 39.3 m/s
a = -9.81 m/s²
t =
Y =
V_y =

But that's all I got.

 

[tms]

All you've done is write down one of the initial conditions and the acceleration due to gravity. What equations describe motion in a line? There are several equations that relate position, velocity, acceleration, and time.

 

[mgerman63016]

I'm sorry I don't know I'm thinking the equations we could use for the problem would be

V_y=V_ox+at
y-y₀=1/2(V_y+V_oy)t
2a(y-y₀) = V_y²-V_oy²
y-y₀=V_oyt+1/2at²
x-x₀=V_oxt

these are the equations I'm suppose to use

 

[FeynmanIsCool]

^{pssst.. maybe} displacement X=V_ot(cosθ) displacement Y= V_ot(sinθ)-1/2(gt²)

 

[mgerman63016]

I've never seen those before.

 

[haruspex]

V_y=V_ox+at

No, that's mixing vertical and horizontal velocities.

V_oy = 39.3 m/s

No, |V_o| = 39.3 m/s. Given the launch angle, how do you determine V_oy and V_ox ?
Once you know V_oy, which equation will you use to find V_y at time t?

 

[voko]

You need equations that

1. Relate velocity components (horizontal and vertical) with the magnitude of velocity (speed) and its direction (angle to the ground), and vice versa.

2. Describe how the vertical velocity component is affected by acceleration due to gravity.

 

[FeynmanIsCool]

I've never seen those before.

Remember, your question has an angle involved, but none of the formulas that you stated included θ as a variable...

 

[FeynmanIsCool]

also remember, horizontal (movement along x) and vertical (movement along y) do not effect each other in ANY way. So, they must be calculate separately. In order to find direction at time (t) you must calculate x and y movement independent of each other

 

[mgerman63016]

to determine Voy and Vox and since I have a angle could I use the cos sin and tan and the pythagorean theorem to determine the distance traveled?

 

[haruspex]

You use sin/cos/Pythagoras to obtain Voy and Vox. Thereafter, treat the vertical and horizontal motions separately to find later velocities and or distances. In this question, you'll need to recombine the velocities later.

 

[mgerman63016]

V_y= 39.3sin(56.2)
V_y= 32.66 m/s

V_x=39.3 m/s cos(56.2)
V_x=21.86

this is what I got so far. I'm trying to work on the Pythagoras theorem right now. But I'm not sure what I'm suppose to use

 

[haruspex]

V_y= 39.3sin(56.2)
V_y= 32.66 m/s

V_x=39.3 m/s cos(56.2)
V_x=21.86

this is what I got so far.

Those are the initial vertical and horizontal velocities. Now you need to use your kinematic equations to find the velocities at the give times.

 

[FaraDazed]

V_y= 39.3sin(56.2)
V_y= 32.66 m/s

V_x=39.3 m/s cos(56.2)
V_x=21.86

this is what I got so far. I'm trying to work on the Pythagoras theorem right now. But I'm not sure what I'm suppose to use

Just remember that the horizontal initial velocity stays constant throughout the motion, so its simply distance=velocity*time. Plus gravity is acting down, in the y direction, so when dealing with the x values, a/g is always 0 (that is why it is just x=v0*t because x=v0t+1/2at^2 is x=v0t+0)

Use the equations of motion you wrote down but when dealing with intial velocity use 39.3cos56.2 (v0cosΘ).