A ball was hit over a net
Vy= 0
Soy= 2.37m
g= -9.8
Vox= 34.42m/s
t= 2.137
Sx= 23.92m
I know the equations used are
Sx= (34.42m/s)t
Sy= 2.37m + 1/2(-9.8)t^2
using:
Sx= Soy + Vxt
Sy= Soy + Vyt + 0.5gt^2
if I'm not incorrect...I plugged in the values and made 2 displacement graphs with respect to time.
How would I go and make a velocity vs time graph for the x component and y component??
Do I go and get the derivative of the eequation or something?
I would just take the derivative of each, and plot them on the same graph (different colors?) vs time.
Vx = Vox at all times
Vy = Voy - gt, but in your case Voy = 0, so
Vy = -gt
(until the ball hits the ground)
To create a velocity vs. time graph for the x and y components, you will indeed need to find the derivatives of the displacement equations with respect to time.
Let's start with finding the velocity in the x-component (Vx). We can calculate this by taking the derivative of the displacement equation with respect to time:
Sy = Soy + Vyt + 0.5gt^2
Differentiating both sides with respect to time, we get:
0 = Vy + gt
Since Vy is given as 0, this simplifies to:
0 = 0 + gt
Therefore, gt = 0. This means the velocity in the y-direction remains constant throughout the motion.
Now, let's find the velocity in the x-component (Vx). The displacement equation for the x-component is given as:
Sx = Vox * t
Differentiating with respect to time, we get:
Vx = d(Sx)/dt = d(Vox * t)/dt = Vox
Hence, the x-component velocity (Vx) remains constant at the value of Vox, which is 34.42 m/s.
To create the velocity vs. time graph for each component, plot a horizontal line at the value of Vx (34.42 m/s) for all time values. This will represent the x-component velocity graph.
For the y-component, since Vy is always 0, the velocity vs. time graph will be a flat line at 0 m/s for all time values.
In summary, the velocity vs. time graph for the x-component will be a horizontal line at 34.42 m/s, and the velocity vs. time graph for the y-component will be a horizontal line at 0 m/s.