Question

A bullet leaves the muzzle of a gun at 250 m/s to hit a target 120m away at the level of the muzzle (1.9 m above ground), the gun must be aimed above the target. (Ignore air resistance.) How far above the target is that point?

Answer 1

The distance is `1.11m`.

Explanation:

To answer at the question we have to find the angle `alpha` above the horizontal, at which we have to shoot.

The motion is a parabolic motion, that is the composition of two motion:

the first, horizontal, is an uniform motion with law:

`x=x_0+v_(0x)t`

and the second is a decelerated motion with law:

`y=y_0+v_(0y)t+1/2g t^2`,

where:

• `(x,y)` is the position at the time `t`;
• `(x_0,y_0)` is the initial position;
• `(v_(0x),v_(0y))` are the components of the initial velocity, that are, for the trigonometry laws:
  `v_(0x)=v_0cosalpha`
  `v_(0y)=v_0sinalpha`
  (`alpha` is the angle that the vector velocity forms with the horizontal);
• `t` is time;
• `g` is gravity acceleration.

To obtain the equation of the motion, a parabola, we have to solve the system between the two equation written above.

`x=x_0+v_(0x)t`
`y=y_0+v_(0y)t+1/2g t^2`.

Let's find `t` from the first equation and let's substitue in the second:

`t=(x-x_0)/v_(0x)`

`y=y_0+v_(0y)(x-x_0)/v_(0x)-1/2g*(x-x_0)^2/v_(0x)^2` or:

`y=y_0+v_0sinalpha(x-x_0)/(v_0cosalpha)-1/2g*(x-x_0)^2/(v_0^2cos^2alpha)` or

`y=y_0+sinalpha(x-x_0)/cosalpha-1/2g*(x-x_0)^2/(v_0^2cos^2alpha)`

To find the range we can assume:

`(x_0,y_0)` is the origin `(0,0)`, and the point in which it falls has coordinates: `(0,x)` (`x` is the range!), so:

`0=0+sinalpha*(x-0)/cosalpha-1/2g(x-0)^2/(v_0^2cos^2alpha)rArr`

`x*sinalpha/cosalpha-g/(2v_0^2cos^2alpha)x^2=0rArr`

`x(sinalpha/cosalpha-g/(2v_0^2cos^2alpha)x)=0`

`x=0` is one solution (the initial point!)

`x=(2sinalphacosalphav_0^2)/g=(v_0^2sin2alpha)/g`

(using the double-angle formula of sinus).

Now let's find, finally `alpha`.

`sin2alpha=(gx)/v_0^2=(9.8*120)/250^2=0.0188°rArr`

`2alpha_1=arcsin0.0188rArr2alpha=1,078°rArralpha=0.53°` and

`2alpha_2=180°-arcsin0.0188=178.92°rArralpha=89.47°`.

The solution are both valid, but the most reasonable is the first.

Using trigonometry, now, we can find the answer.
There is a right-angled triangle with an acute angle of `0.53°`, and a cathetus of `120m`. The other cathetus is

`c=120*tan0.53=1.11m`.