Kinamatics problem please help <3

[physicsprincezz88]

1. A person shoots off a shotgun so that we know the muzzle velocity is 2,200 ft / sec. If that shooter is standing in such a way that the gun is angled at 15*degrees above a horizontal line 5 feet off the ground. It is given that the vertical distance can be calculated by the equation y = -G/2 t [sup:il9y1vrs]2[/sup:il9y1vrs] + V[sub:il9y1vrs]o[/sub:il9y1vrs] Sin (~)t + y[sub:il9y1vrs]o[/sub:il9y1vrs] and the horizontal distance is x=V[sub:il9y1vrs]o[/sub:il9y1vrs] Cos (~) t



Relevant equations :

y = -G/2 t [sup:il9y1vrs]2[/sup:il9y1vrs] + Vo Sin (~)t + Y[sub:il9y1vrs]o[/sub:il9y1vrs]
x=V[sub:il9y1vrs]o[/sub:il9y1vrs] Cos (~)t

Where the (~) is a theta and the o after the V and Y mean initial.



Questions.
What is the equation vertical and horizontal distances for the shooter?

(A) What time does the projectile reach max height?
(B) What is the max height?
(C) What is the flight time?
(D) How far does the projectile fly?
(E) Find an equation that models the flight path (hint: y in terms of x)
(F) What is the velocity when it hits the ground?
(G) If the 12 ft wall is put 50 ft away does the projectile hit the wall?
(H) If it does what is the velocity when it hits the wall?





Thanks so much, if anything I'm not even asking for answers but how I get them.

I have been struggling with physics homework and this is one of the ones that give me so much trouble...

 

[galactus]

1. A person shoots off a shotgun so that we know the muzzle velocity is 2,200 ft / sec. If that shooter is standing in such a way that the gun is angled at 15*degrees above a horizontal line 5 feet off the ground. It is given that the vertical distance can be calculated by the equation \(\displaystyle y = \frac{-g}{2}t^{2} + V_{0} Sin (\theta)t + y_{0}\) and the horizontal distance is \(\displaystyle x=V_{0} Cos (\theta)t\)

Here's hints on the first few.

What is the equation vertical and horizontal distances for the shooter?

\(\displaystyle x(t)=(2200cos(15))t, \;\ y(t)=5+(2200sin(15))t-16t^{2}\)

(A) What time does the projectile reach max height?

The max height occurs when \(\displaystyle \frac{dy}{dt}=0\)

Find the derivative of y(t), set to 0 and solve for t.

Or, use \(\displaystyle t=\frac{-b}{2a}, \;\ y=c-\frac{b^{2}}{4a}\)

(B) What is the max height?

Sub the value found for t from part a into y.

(C) What is the flight time?

Set y(t)=0 and solve for t. There will be two values: The initial time and the impact time.

(D) How far does the projectile fly?

Sub in the value found in part c into x(t).

(E) Find an equation that models the flight path (hint: y in terms of x)

Solve x(t) for t and sub into y(t). It will then be y in terms of x.

For part f, a vector function, r(t), can be used as well for these problems. For this portion, differentiate to get v(t), and sub in the time to impact from part c.

\(\displaystyle r(t)=550\sqrt{2}(\sqrt{3}+1)t\cdot i+[(-16t^{2}+550\sqrt{2}(\sqrt{3}+1)t+5]\cdot j\)

Then, find the norm \(\displaystyle ||v(t)||\) to give speed at impact.

 

[physicsprincezz88]

Galactus.. Thank you

Attempting as we speak.

I'll write back tonight and tell you how it goes.