Setting up and solving IVPs for projectile motion

[jwpaine]

A projectile is launched off of a 20 foot high wall at an angle of 25 degrees and a velocity of 30ft/sec

I'm leaving a few details out, but basically I need to find how far away from the wall the projectile lands at y = 0

Breaking the initial velocity into vertical and horizontal components, I have Vo upwards = 30sin(25) and Vo horizontal = 30cos(25)
Gravity = -32ft/s^2
So for position at time t, x(t) = 30cos(25)t and y(t) = 30sin(25)t - 32t^2 + 20
Solving for 0 = 30sin(25)t - 32t^2 + 20 I get t approx = 1 (we only care about a positive time here)
substituting t = 1 back into x(t) I get how far the projectile went, horizontally, before landing.

The above is how I would have done it in high school physics, but I'm supposed to setup and solve IVPs for x and y.
So my initial conditions are:

x''(t) = 0, x'(0) = 30cos(25), x(0) = 0
y''(t) = -32ft/sec^2, y'(0) = 30sin(25), y(0) = 20feet

I'm not sure what I should do from here? I know I will have to integrate, such like int(x''(t)) = C = 30cos(25)
Can someone give me a hand please?

Thanks!

 

[TchrWill]

A projectile is launched off of a 20 foot high wall at an angle of 25 degrees and a velocity of 30ft/sec

I'm leaving a few details out, but basically I need to find how far away from the wall the projectile lands at y = 0

Breaking the initial velocity into vertical and horizontal components, I have Vo upwards = 30sin(25) and Vo horizontal = 30cos(25)
Gravity = -32ft/s^2
So for position at time t, x(t) = 30cos(25)t and y(t) = 30sin(25)t - 32t^2 + 20
Solving for 0 = 30sin(25)t - 32t^2 + 20 I get t approx = 1 (we only care about a positive time here)
substituting t = 1 back into x(t) I get how far the projectile went, horizontally, before landing.

The above is how I would have done it in high school physics, but I'm supposed to setup and solve IVPs for x and y.
So my initial conditions are:

x''(t) = 0, x'(0) = 30cos(25), x(0) = 0
y''(t) = -32ft/sec^2, y'(0) = 30sin(25), y(0) = 20feet

I'm not sure what I should do from here? I know I will have to integrate, such like int(x''(t)) = C = 30cos(25)
Can someone give me a hand please?

IN FACT:
Breaking the initial velocity into vertical and horizontal components, yields Vo upwards = 30sin(25) = 12.678fps. and Vo horizontal = 30cos(25) = 27.189fps.
Gravity = -32ft/s^2
Therefore, for position at time t,
x(t) = 30cos(25)T
t(up) to peak height is 30sin25/32 = .396sec.
t(dwn) = t(u) to starting point where V(dwn)=30sin25=12.678fps.
y(t) from starting point = 20 = 12.678t + 32t^2/2 or
16t^2 + 12.578t -20 = 0
Solving,t = .79sec. (we only care about a positive time here)
Total flight time T = 2(.396) + .79 = 1.582sec.
Substituting T = 1.582 back into x(t), we get how far the projectile went, horizontally, x(t) = 30cos25(1.582)= 43ft.

I'll leave the IVP's to someone else.

Hope I did not miss anything.

 

[jwpaine]

A projectile is launched off of a 20 foot high wall at an angle of 25 degrees and a velocity of 30ft/sec

I'm leaving a few details out, but basically I need to find how far away from the wall the projectile lands at y = 0

Breaking the initial velocity into vertical and horizontal components, I have Vo upwards = 30sin(25) and Vo horizontal = 30cos(25)
Gravity = -32ft/s^2
So for position at time t, x(t) = 30cos(25)t and y(t) = 30sin(25)t - 32t^2 + 20
Solving for 0 = 30sin(25)t - 32t^2 + 20 I get t approx = 1 (we only care about a positive time here)
substituting t = 1 back into x(t) I get how far the projectile went, horizontally, before landing.

The above is how I would have done it in high school physics, but I'm supposed to setup and solve IVPs for x and y.
So my initial conditions are:

x''(t) = 0, x'(0) = 30cos(25), x(0) = 0
y''(t) = -32ft/sec^2, y'(0) = 30sin(25), y(0) = 20feet

I'm not sure what I should do from here? I know I will have to integrate, such like int(x''(t)) = C = 30cos(25)
Can someone give me a hand please?

IN FACT:
Breaking the initial velocity into vertical and horizontal components, yields Vo upwards = 30sin(25) = 12.678fps. and Vo horizontal = 30cos(25) = 27.189fps.
Gravity = -32ft/s^2
Therefore, for position at time t,
x(t) = 30cos(25)T
t(up) to peak height is 30sin25/32 = .396sec.
t(dwn) = t(u) to starting point where V(dwn)=30sin25=12.678fps.
y(t) from starting point = 20 = 12.678t + 32t^2/2 or
16t^2 + 12.578t -20 = 0
Solving,t = .79sec. (we only care about a positive time here)
Total flight time T = 2(.396) + .79 = 1.582sec.
Substituting T = 1.582 back into x(t), we get how far the projectile went, horizontally, x(t) = 30cos25(1.582)= 43ft.

I'll leave the IVP's to someone else.

Hope I did not miss anything.

Thanks... I knew that x = xo + vot + (1/2)at^2
I am not sure why I forgot and did it that way.... I had to look up the derivation for that kinematic equation.

Now if I can get some help from someone so I can do with IVPs

 

[galactus]

Hey JW:

This is Skeeter's forte, but I think I can help a little.

The acceleration of gravity is a constant. \(\displaystyle a_{y}=\frac{d^{2}y}{dt^{2}}=-32\)..........[1]

The horizonatal acceleration is 0: \(\displaystyle a_{x}=\frac{d^{2}x}{dt^{2}}=0\)...........[1]

Integrate [1] wrt t: \(\displaystyle v_{y}=-\int{32}dt=32t+C_{1}\)

\(\displaystyle y=\int{v_{y}}dy=-16t^{2}+C_{1}t+C_{2}\).......[3]

\(\displaystyle v_{x}=\int{a_{x}}dt=0+C\)

\(\displaystyle x=\int{v_{x}}dt=\int{C}dt\)

\(\displaystyle x=Ct+C_{3}\)

\(\displaystyle C_{3}=0\)

y is 20 when t=0, so if we sub in t=0 in equation [3], we find that \(\displaystyle C_{2}=20\)

The vertical components of velocity is \(\displaystyle v_{y}=30sin(25)=12.68=C_{1}\)

The horizontal component is \(\displaystyle v_{x}=30cos(25)=27.19=C\)

Sub these back into x and y to find the equations we need.

\(\displaystyle x=30cos(25)t\)

\(\displaystyle y=-16t^{2}+30sin(25)t+20\)

There, we derived the components using integration and a few little DE's. Was that the idea?.

 

[jwpaine]

Hey, thanks for the reply - this almost looks too easy, but I guess this is the only thing he could be asking for...hmm
I'm shooting him an email to make sure.

Thanks for the help!