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?
The distance is
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:
#(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:
#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.
To find the range we can assume:
(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