why is that the range of the projectile launched at 30degree to the one launched at 60 is the same?

Question

why is that the range of the projectile launched at 30degree to the one launched at 60 is the same?can you expalin to me why it is the same...it has a theory???please state to me

and why is not the same ,that the maximum vertical distance reached by the projectile launched at 30 degree to the one launched at 60 degree? ?

many thanks

Answer 1

Assuming no wind resistance, the optimum trajectory of a projectile will always be at an angle of 45 degrees.

A projectile's initial velocity drives the rest of its trajectory. That velocity can be broken into two parts -- one horizontal and one vertical. The time the projectile stays in the air is determined by the vertical velocity. The faster it takes off vertically, the higher it goes up, which means it will stay in the air longer. If you multiply that time by the horizontal velocity, you see how long the projectile has traveled horizontally. Thus, there is a pressure to increase the vertical velocity in order to increase the time in the air AND to increase the horizontal velocity to increase the speed traveling horizontally.

It turns out that the best balance of these things is a 45 degree angle. This makes the initial velocity split equally between the horizontal and the vertical. 60 degree is 15 too high, so there's too little horizontal velocity to make it travel far, and 30 degrees is 15 too low, so there's too little vertical velocity to keep it in the air.

Note that this is also why a 60 degree angle comes with a higher maximum vertical distance than a 30 degree angle. It travels much farther up than the 30 degree angle because it's directed into the sky more. However, it travels much more slowly in the horizontal direction, which is why it doesn't travel as far as the 30 degree angle. In other words, the 60 degree angle makes a very HIGH arch. The 30 degree angle makes a very SHALLOW arch because not much of its speed is directed vertically, thus it will hit the ground sooner.

If you want proof of this, take the expression 0.5*a*t^2 + v0*t + x0. This is the expression for the displacement of a projectile given an initial velocity of v0, an initial distance of x0, and an accleration of a. If we replace a with (-g), x0 with 0, and v0 with the initial VERTICAL velocity, this becomes the vertical displacement at any time:

v*sin(angle)*t - 0.5*g*t^2

Notice that I replaced v0 with v*sin(angle). This is the component of the velocity directed in the vertical direction. I want to solve for the time when this expression is 0. Clearly, it is zero at t=0 because I started on the ground initially, so I'm going to divide out the t=0 and solve for the remaining root. Thus, the object hits the ground at:

t = v*sin(angle)/(0.5*g)

As I said, the total HORIZONTAL displacement is simply this time multiplied by the horizontal velocity, which is just v*cos(theta). So the total HORIZONTAL distance traveled is:

v*cos(angle)*t = v * cos(angle) * v * sin(angle) / (0.5*g)
= v^2 * cos(angle) * sin(angle) / (0.5*g)

Our goal is to maximize this expression with respect to the angle. Since v^2/(0.5*g) is positive, then this is equivalent to maximizing cos(angle)*sin(angle). To do this, take the derivative of cos(angle)*sin(angle) with respect to angle. You get:

cos(angle)*cos(angle) - sin(angle)*sin(angle)

I want to solve for the angle that makes this zero. In other words, I want to solve for where:

cos^2(angle) = sin^2(angle)

But, sin^2(angle)=1-cos^2(angle) (by Pythagoras theorem), so

cos^2(angle) = 1 - cos^2(angle)

Which means:

cos^2(angle) = 1/2

So I want to solve for the angle between 0 and 90 degrees where:

cos(angle) = 1/sqrt(2)

This angle is 45 degrees. If I take the second derivative of the expression I started with and evaluate it at 45 degrees, I get a negative number at 45 degrees (-2), thus I know that a 45 degree angle is a maximum. Of course, I also need to plug in 0 degrees and 90 degrees into my original expression to verify that they don't beat 45 degrees, and it turns out that they do not either.

So the optimal angle is 45 degrees.

Answer 2

That's a very interesting question.

Let's now prove that the range R for the projectile launched at 30 degrees is the same as that at 60 degrees.

Let's first consider the one at 30 degrees where the speed of the projectile at launch is x:

The horizontal component of the speed is xcos30, where x is the initial speed. R = speed multiplied by t. or xcos30*t.

Let's compute for t:

v=u+at
0=xsin30+(-9.8)t (v=0 because that's the speed at the top of the flight of the projectile. u is the vertical component of the initial speed x. 9.8 is negative because the projectile is moving opposite the direction of gravity.)

t=xsin30/9.8 . This is the time to reach the top. The total time is twice this or 2xsin30/9.8

Substitute in our formula for R:

R=xcos30*2xsin30/9.8

Now the one at 60 degrees:

The horizontal component of the speed is xcos60, using the same formula as above R=xcos60*t

Let's compute also for t:

v=u +at
0=xsin60+(-9.8)t
t=xsin60/9.8 . this again is just half of the time it takes for the projectile to reach the ground. So we also multiply by 2. Thus total t=2xsin60/9.8

Substitute in our formula for R:

R=xcos60*2xsin60/9.8

Now we have to prove that the R for 30 degrees
equals the R for 60 degrees:

xcos30*2xsin30/9.8=xcos60*2sin60/9.8

This takes a little knowledge of trigonometry. Draw a 30-60 right triangle. Observe. Isn't sin30
equal to cos60 and cos30 equal to sin60? Now
substitute in the left hand side of our equation:

xsin60*2xcos60/9.8=xcos60*2sin60/9.8

or rearrange the left hand side to make it clearer:

xcos60*2xsin60/9.8=the same as the right hand side.

Next, why is the maximum vertical distance reached by the projectile launched at 30 degrees not the same as that at 60 degrees.

Use the formula v^2-u^2=2as, where v = 0 again. Now solve for s.

s=u^2/2a

The vertical component of the initial speed as we had shown above is xsin30 for the one launched at 30 degrees and xsin60 for the one launched at 60 degrees. Here the story is different because sin30=0.5, and sin60=0.866. Substituting these values in the above equation , we will of course get a different value for s at 30 degrees, and for s at 60 degrees.

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questions in Physics today.

Answer 3

Because maximum distance will be achieved with a trajectory of 45 degrees. These two values are off from the optimum by the same amount (15 too high and too low) So at 60 degrees you will get greater vertical distance, but not enough horizontal- while at 30 degress gravity will bring the projectile to earth faster because it has a shorter vertical distance to fall.

Answer 4

lets assume Vo is the initial velocity and @ is the angle , Vx is the velocity according to axis X and Vy is the velocity according axis y, g is gravity and t is time.
Then Voy=Vo sin @ and Vox = Vo cos @
Vx= Vox = Vo cos @
Vy= Voy-g.t = Vo sin @ - g t
time needed to go to max vertical distance is when Vy=0 that is Vo sin @ - gt = 0 hence t=(Vo sin @)/g
and time to go back to original level is 2 t = 2(Vo sin @)/g
And we get the max distance at horizontal level is =X= Vx . 2t = Vo cos @ . 2t = Vo cos @ . 2(Vo sin @)/g
equals 2 Vo^2 sin @ cos @ /g = Vo^2 . sin 2@ /g
when you put @ = 30 you will find the same X as you put @ = 60
the same way you will find the distance Y=Vo sin @ . t - g.t/2 is different for @=30 compared with @=60

Answer 5

the range depends on sin(2*angle)....so the ratio is 1....
the max height is proportional to sin^2