How to find the horizontal distance from projectile motion?

[Eclair_de_XII]

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.

$x-x_0=(v_0cosθ)t$
$x=(250\frac{m}{s})(cos(-6.823°))(3.03s)=752m≠758m$

 

[berkeman]

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.

$x-x_0=(v_0cosθ)t$
$x=(250\frac{m}{s})(cos(-6.823°))(3.03s)=752m≠758m$

I'm not following your calculations. The vertical motion is the same as if the projectile were dropped from that same height. How long does it take an object to fall that distance if dropped from rest?

And given the horizontal velocity and that time, how far does the projectile go before it hits the ground.

And how fast is an object going vertically when dropped from that height and it hits the ground?

 

[Eclair_de_XII]

I'm not following your calculations.

I make the final velocity (when the bullet hits the ground) to be zero.

 

[berkeman]

I make the final velocity (when the bullet hits the ground) to be zero.

ROTFLMAO!

 

[Eclair_de_XII]

Uh... Okay, so what don't you understand about my calculations?

 

[SteamKing]

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.

$x-x_0=(v_0cosθ)t$
$x=(250\frac{m}{s})(cos(-6.823°))(3.03s)=752m≠758m$

The reason your range value differs slightly from the book value is that you have calculated the range assuming the gun is pointed below the horizontal. That's what the factor cos (-6.823°) would indicate. The angle of firing, θ₀ ≠ θ on impact.

The gun is fired horizontally, according to the problem statement, so the projectile is traveling horizontally as it leaves the gun. This means the angle θ₀ = 0°, and the initial vertical velocity of the projectile v₀ ⋅ sin θ₀ = 0 as it leaves the gun. This also implies that the horizontal velocity is v₀ ⋅ cos θ₀ = v₀ = 250 m/s, as specified.

The projectile stays in the air only as long as it takes it to free fall from the elevation of 45 m.

 

[berkeman]

I make the final velocity (when the bullet hits the ground) to be zero.

So if I drop you off a 45m building, you will be going 0m/s when you reach the ground and will not be hurt. And I will not be in trouble with the Physics Police...

 

[cnh1995]

I make the final velocity (when the bullet hits the ground) to be zero.

Is it really zero?

 

[Eclair_de_XII]

That's what the factor cos (-6.823°) would indicate. The angle of firing, θ_0 ≠ θ on impact.

The gun is fired horizontally, according to the problem statement, so the projectile is traveling horizontally as it leaves the gun. This means the angle θ0 = 0°, and the initial vertical velocity of the projectile v0 ⋅ sin θ0 = 0 as it leaves the gun. This also implies that the horizontal velocity is v0 ⋅ cos θ0 = v0 = 250 m/s, as specified.

Oh, so it's just $(250\frac{m}{s})(cos0°)(3.03s)=757.5m$

So if I drop you off a 45m building, you will be going 0m/s when you reach the ground and will not be hurt.

Yeah... I guess you have a point. Still, though, the book says that the velocity just before the bullet reaches the ground is equal to $29.7\frac{m}{s}$, which is what I calculated. But then that's when I set final velocity to zero... I guess that the initial velocity is the one that's zero? Since horizontal motion doesn't affect vertical motion, and the vertical velocity would be just the same as if there were no horizontal motion at all. Anyway, I think I have my answers. It asks for the magnitude of the final vertical velocity, which is ##|-29.7\frac{m}{s}|=29.7\frac{m}{s}. Anyone feel free to correct me.

 

[berkeman]

I guess that the initial velocity is the one that's zero?

Correct. Problems like this are just about separating the vertical and horizontal motion. If the initial launch angle were different from zero from the horizontal, you would use trig to figure out the vertical component of velocity. As it is, it's just a simple drop problem with a constant horizontal velocity.