A bullet hole is 80" (inches) from the floor against a wall.
A shoe print from the shooter is 45" away from the wall.
The bullet was shot from a 65 degree angle.
What is the shooter's shoulder height? (from where he fired the gun)
The shooters height will be (80 + x) inches
The third angle in the triangle is 180 - 90 - 65 = 25º
tan25º = x/45
45tan25º = x
Shooter's height = 80 + 45tan25 ≈ 100.984 inches
That is over 8 feet tall!
Okay I just figured out that it probably means this:
In which case the height is 80 - 45tan25 ≈ 59.016 inches
About 4.9 feet, which makes more sense.
Hey hectictar, thanks for responding.
I may have explained this poorly. The shooter is not 100 inches tall, heh.
The wall is X inches tall. The bullet hole is in the wall at 80 inches up.
The shooter shot from 45 inches away.
He shot UPWARDS at 65 degrees. It's not like he was pointing the gun straight at the wall.
I cannot get these numbers to work...... are you sure you have them correct???
From 45 inches away a 65 degree angle would put the bullet at
tan 65 = opp/adj = opp/45
45 tan 65 = 96 inches ABOVE shoulder height and the hole is only 80 inches above the floor....which would mean his shoulder was 16 inches BELOW the floor.....
If we assume the bullet travels in a straight line (it doesn't) we do not need to know the muzzle velocity.
Let's assume the shot is UPWARD at a 6. 5 degree angle
tan 6.5 = opp/45 opp = 5.127 "
80-5.127 = 74 inch shoulder height (pretty tall people are usually about 7-8 heads tall ...this person would be ~~86 inches tall ....IF the angle is 6.5 degrees upward)
I actually drew the second picture wrong. I think the confusion is over whether the 65º angle is from vertical or horizontal. This is how the second one I did should look like:
I was making it really fast so I messed up.
But when it's like this the height actually is 80 - 45tan25º ≈ 59.016 inches