physics projectile motion help plzzzzz its tough?

Question

prove that a gun will shoot three times as high when its angle of elevation is 60 degrees as when it is 30 degrees, but will carry the same horizontal distance in each case.

Answer 1

When you fire a gun at an angle, you can express its velocity as a horizontal component and a vertical component.

Let's say that the initial velocity is x. The horizontal component is cos(x) and the vertical component is sin(x). Then when the gun is aimed at 60 degrees, the horizontal component is .5 and the vertical is .86(x) [sqrt(.75)]. When the gun is aimed at 30 degrees, the two are reversed.

Now, the velocity would remain constant, except that there is a vertical acceleration due to gravity, which will be expressed as a negative number because it is acting against the velocity. These days, this is said to be 10m/s^2. v=at, so t=v/a, so at 60 degrees it takes t=(.5xm/s)/(10m/s^2) or .05s to reach 0 velocity. It then takes another .05xs for the object to fall to earth, for a total of .1xs.

Gravity doesn't act on the horizontal motion, which will continue until the vertical motion causes it to impact with the Earth, at .1s. So at 60 degrees, d=vt = .86xm/s * .1s = .086xm. Remember that x is the initial velocity. So if x were 20 m/s, the distance would be .086 * 20m = .172m.

Now let's do it at 30 degrees.

t = v/a = .86xm/s / 10m/s^2 = .086s to zero vertical velocity = .172s to impact.

d = vt = .5xm/s & .172s = .086xm

Now, why does it work.

Look at the two equations.

d=v^2/a

At 60 degrees: d=.5xm/s * (.86xm/s / 10m/s^2)
At 30 degrees: d=.86xm/s * (.5xm/s / 10m/s^2)

Answer 2

Calm down. These are easily doable things if you know the equations. Your problem is simply that no one gave you the equations. Here they are:
To calculate the range (horizontal distance) of any projectile use this equation:
Range = initial velocity squared times the sine of twice the angle divided by the acceleration due to gravity (g) or R = vi^2 sin 2theta / g. (Note that the initial velocity is the velocity at the angle of the projectile’s path from the horizontal.)

Take 30 and 60 degrees as your two angles. (note that the range is the same for angles that equal 90. For example the horizontal range of 70 and 20 are the same, as are 10 and 80) The sine of 120 (60 x 2) and the 60 (30 x 2) are the same. Thus the horizontal distance is the same for these two angles.

As for the vertical distance you can use this equation: distance vertical = the square of the product of the vertical velocity and the sine of the angle divided by g. Thus dv = (vv x sine theta)^2 / g. This works because the square of sine 30 degrees is 1/3 of the square of sine 60 degrees (Sine 30 = 0.5; 0.5^2 = 0.25. Sine 60 = 0.866; 0.866^2 = 0.75)

I hope this helps!

Answer 3

Alright, I didn't really work this out, but the only way you'd be able to prove that is to break down the velocity into two separate vectors (the one moving upwards, and the one moving horizontal) for each situation (this can be done with some manipulation of theta and triangle rules from algebra) and find how far each situation would make the projectile go in either plain.

Answer 4

Let the gun be shot with a velocity of x m/s each time.

1st time:
Horizontal speed = x cos 30 = [sqrt(3)*x]/2.
Vertical speed = x sin 30 = x/2.

The acceleration in the vertical direction on the gun is -9.81m/s^2.

To reach maximum vertical speed when it's vertical speed is zero:
Time taken = [0-x/2] / -9.81 = x/19.62 s.

Horizontal direction travelled in this time:
= [sqrt(3)*x]/2 * x/19.62
= [sqrt(3)*x^2] / 39.24 m.

Vertical direction travelled in this time:
= (v+u)t / 2
= (x/2)/2 * x/19.62
= x^2 / 78.48 m.


2nd time:
Horizontal speed = x cos 60 = x/2
Vertical speed = x sin 60 = [sqrt(3)*x]/2.

The acceleration in the vertical direction on the gun is -9.81m/s^2.

To reach maximum vertical speed when it's vertical speed is zero:
Time taken = [0-[sqrt(3)*x]/2] / -9.81
= [sqrt(3)*x] / 19.62 s.

Horizontal direction travelled in this time:
= x/2 * {[sqrt(3)*x] / 19.62}
= [sqrt(3)*x^2] / 39.24 m.

Vertical direction travelled in this time:
= (v+u)t / 2
= ([sqrt(3)*x]/2)/2 * [sqrt(3)*x] /19.62
= 3x^2 / 78.48 m.

As we can see, the horizontal distance travelled in both instances
= [sqrt(3)*x^2] / 39.24 m.

The vertical distance travelled when angle is 60 degrees = 3 * vertical distance travelled when angle is 30 degrees
= 3 * x^2 / 78.48 m
= 3x^2 / 78.48 m.

Answer 5

most ppl would say that it would be twice as high but because of the distance between the barrel and the target it will shoot higher than expected