Projectile Horizontal Range: Debunking the Intuition

[yitriana]

The equation describing the horizontal range of a projectile is

[tex]
R = \frac{v_0^2}{g} \sin{2\theta}
[/tex]

The previous equation states that the projectile wouldn't travel any distance if launched at 0 degrees. Further, the range of a projectile decreases from 45 to 0 degrees, but it seems counterintuitive, since the horizontal distance traveled is described by,

[tex]
x = v_0 \cos{\theta_0}t
[/tex]

Therefore, an angle of 0 would yield maximum x value.

I'm rather confused.

 

[tiny-tim]

Hi yitriana!

(have a theta: θ and a degree: º )

… Further, the range of a projectile decreases from 45 to 0 degrees, but it seems counterintuitive, since the maximum distance traveled is when,

[tex]
x = v_0 \cos{\theta_0}t
[/tex]

Therefore, an angle of 0 would yield maximum x value.

Yes, the range is v₀cosθ times t,

and cosθ is maximum when θ = 0,

but t is the time in the air, and it isn't constant.

sin2θ is maximum when 2θ = 90º, ie θ = 45º.

 

[Bob S]

Write out the equations of motion for both v_x and v_y at time t=0 for initial velocity v₀.

v_x0 = v₀ cosθ
v_y0 = v₀ sinθ

The maximum height of the projectile is given by
mgh = (1/2)mv_y0²; solve for h
the time the projectile is in the air is given by
t=2 sqrt(2 h/g)
The range is
R=v_x0 t

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω

 

[yitriana]

Yes, the range is v₀cosθ times t,

and cosθ is maximum when θ = 0,

but t is the time in the air, and it isn't constant.

sin2θ is maximum when 2θ = 90º, ie θ = 45º.

Yes, I see that R is maximized at θ = 45º

However, if we were to find the x distance of the projectile, wouldn't the distance be maximized when θ= 0, since v_x0 = v₀ cosθ is greatest?

How are the x distance and range different when θ = 0º?

I know they are different when θ != 0º

It seems logical that if the projectile goes in a straight line with constant velocity, it will travel farther than if the projectile goes in a curved path (at an angle) at constant x velocity.

So how do the two equations differ when launching a projectile at 0º??

 

[Cantab Morgan]

The assumption is that friction with the ground will halt the horizontally fired projectile immediately. In other words, [itex]R = 0[/itex] when theta is zero, no matter what the initial velocity is. If your cannon is actually some vertical distance above the ground when it fires, then the formula for range needs some modification. Geometrically, you're looking for the intersection of a line (the ground) and a parabola (the trajectory).

 

[protonchain]

The equation describing the horizontal range of a projectile is

[tex]
R = \frac{v_0^2}{g} \sin{2\theta}
[/tex]

The previous equation states that the projectile wouldn't travel any distance if launched at 0 degrees. Further, the range of a projectile decreases from 45 to 0 degrees, but it seems counterintuitive, since the horizontal distance traveled is described by,

[tex]
x = v_0 \cos{\theta_0}t
[/tex]

Therefore, an angle of 0 would yield maximum x value.

I'm rather confused.

Another way to think about it is if [tex] \theta = 0 [/tex] then

[tex] t = \frac{vsin\theta}{g} = \frac{vsin(0)}{g} = 0 [/tex]

since the time of flight is 0, then the distance x is 0.

Btw [tex] t = \frac{vsin\theta}{g} [/tex] comes from doing [tex] v_f = v_0 + gt [/tex] for the y direction, from t = 0, to t = t/2 (aka halfway through). This means [tex] v_f = 0 [/tex] since the ball stops moving in the y direction at that point before starting to come down.

 

[yitriana]

Thanks for all your replies!