- #1

Eclair_de_XII

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## Homework Statement

"A projectile is fired horizontally from a gun that is 45.0 m above flat ground, emerging from the gun with a speed of 250 m/s. (a) How long does the projectile remain in the air?

**(b) At what horizontal distance from the firing point does it strike the ground?**(c) What is the magnitude of the vertical component of its velocity as it strikes the ground?"

## Homework Equations

##v_y^2=(v_0sinθ_0)^2-2g(y-y_0)##

##v_y=v_0sinθ-gt##

##x-x_0=(v_0cosθ_0)t##

##x_0=0m##

##y_0=45m##

##y=0m##

##v_0=250\frac{m}{s}##

__Answer as given by the book:__

##x = 758 m##

## The Attempt at a Solution

I make gravitational acceleration to be positive, because as it hits the ground, the bullet is gaining speed; not losing.

##0=[(250\frac{m}{s})(sinθ)]^2+2(9.8\frac{m}{s^2})(0m-45m)##

##882\frac{m^2}{s^2}=(62500\frac{m^2}{s^2})(sin^2θ)##

##sinθ=-0.1188##

I make the value of sinθ to be negative, because I'm making the gun to face the rightward direction. If it shoots, then it goes down, from the path driven by its velocity.

##θ=-6.823°##

##v_y=v_0sinθ-gt##

##0=(250\frac{m}{s})(sin(-6.823°))+(9.8\frac{m}{s^2})(t)##

##-(9.8\frac{m}{s^2})(t)=(-29.7\frac{m}{s})##

##t=3.03s##

Then here's where I'm having conflicting numbers, when I calculate the horizontal distance.

##x-x_0=(v_0cosθ)t##

##x=(250\frac{m}{s})(cos(-6.823°))(3.03s)=752m≠758m##