What angle yields the maximum horizontal range for a projectile?

In summary, the maximum horizontal range of a plastic dart gun can be found by firing at a 45 degree angle to the horizontal. This is a widely-known concept, as at 90 degrees the range is zero and at 0 degrees the range is also zero. By setting y = 0 and solving for t in the equation of motion, then using this t in the equation for x, we can obtain the range equation which gives x as a function of angle. To find the maximum range, differentiate x with respect to theta, set it equal to zero, and solve for theta.
  • #1
TickleMeElma
16
0
You buy a plastic dart gun, and you wnt to find its max. horizontal range. you shoot the gun straight up, and it takes 4s for the dart to land back at the barrel. What is the maximum horizontal range of your gun?

I got the initial velocity, which is 19.6 m/s. I know I have to incorporate vectors. The initial velocity in the horizontal direction is 19.6m/s (cos x) and in the vertical direction it is 19.6m/s (sin x). Assuming x is the magnitude of the angle. I know that the total displacement in the vertical direction is 0.

And now I am stuck...How do I proceed from here? Please help!

Thanks so very much.
 
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  • #2
Hint: What value for the angle x will produce the maximum range?
 
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  • #3
Your [itex] \vec{v_i} [/itex] vector points at whatever direction you fire at. To fire for maximum distance you want to fire at a 45 degree angle to the horizontal, which means you'll have velocity vector [itex] \vec{v_i} = 19.6m/s @ 45 \ deg [/tex]. Find how far that goes.
 
  • #4
Thank you so much for your help - yeah, somehow, I missed the sample problem that dealt with the apparently widely-known concept of 45 degrees being the angle that yields the max horizontal replacement. Went back and reworked it with your suggestions and those of the book and got ~39m. :)
 
  • #5
apparently widely-known concept of 45 degrees being the angle that yields the max horizontal replacement

It can be sort of intuitive. If the angle is 90 (i.e. straight up), the range is zero. If the angle is zero, the range is zero. Thus, by reasoning, the max would be 1/2 way between the two, or 45 degrees.

Actually, there is a rigorous way to prove it using the equations of motions for a projectile y(t) and x(t). By setting y = 0 and solving for t you get the time it takes to complete the motion. Using this t in the equation for x, you get what's called the range equation, which gives you x as a function of angle. Since you want max x, differentiate x with respect to theta, set it equal to zero, and solve for theta.
 

1. What is "max horizontal displacement"?

"Max horizontal displacement" refers to the maximum distance that an object or point has moved horizontally from its starting position. It is used to measure the amount of displacement in a horizontal direction.

2. How is "max horizontal displacement" different from "vertical displacement"?

"Max horizontal displacement" measures the distance an object has moved horizontally, while "vertical displacement" measures the distance an object has moved vertically. They are two different measurements used to describe the motion of an object in different directions.

3. What factors affect the "max horizontal displacement" of an object?

The "max horizontal displacement" of an object can be affected by the initial velocity, acceleration, and mass of the object. Other factors such as air resistance and friction can also impact the displacement.

4. How is "max horizontal displacement" calculated?

The "max horizontal displacement" can be calculated using the equation: d = v0t + 1/2at^2, where d is the displacement, v0 is the initial velocity, t is the time, and a is the acceleration. This equation is derived from the basic kinematic equations of motion.

5. Why is "max horizontal displacement" important in physics?

"Max horizontal displacement" is important in physics because it is a key measurement used to describe the motion of an object. It helps us understand and predict the behavior of objects in motion, and is crucial in fields such as engineering and mechanics.

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