# 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?

##### 1 Answer

The distance is

#### Explanation:

To answer at the question we have to find the angle

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

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

and the second is a decelerated motion with law:

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.

Let's find

To find the range we can assume:

**is** the range!), so:

(using the double-angle formula of sinus).

Now let's find, **finally**

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