Zizfizziks

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 2 + Vo Sin (~)t + yo and the horizontal distance is x=Vo Cos (~) t

Relevant equations :

y = -G/2 t + Vo Sin (~)t + Yo
x=Vo 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...

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brushman

When the projectile is at its maximum height, it is not moving in the y component. So, if we set the y equation for velocity equal to 0, we can find the time that it is not moving, which is also the time it is at it's maximum height.

Since you are given the distance equation in terms of t, you can find the velocity in terms of t (take the derivative).

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