A long shot

Classical Mechanics Level 3

A marksman is attempting to hit a target that he cannot see, because of an obstacle blocking his view. He knows the target is at the same elevation and $1850$ meters from his location. He also knows that the target is on a vertical wall. With his high power gun site he can see a vertical line extending upward on the wall. He knows the target is on that line. He can see marking or features on the wall that enable him to determine the vertical distance above the target.

What is the vertical distance above the target where he should aim in order to hit the target? Give your answer in meters.

Details and Assumptions

The mean horizontal speed of the bullet is 420 m/s.
The gravitational acceleration is $g=9.8\text{ m/s}^2.$
The gun site is set parallel to the guns barrel for the shot.

The answer is 95.07.

1 solution

Darryl Dennis
Jan 30, 2015

time of flight of the shot $\frac{1850}{420} = 4.4048$ sec

bullet drop do to gravity in 4.405 sec

$\frac{(4.405)\ ^ 2 * 9.8}{ 2} = 95.07$ meter.

You're clever like the Marksman!!! Brilliant Problem!

Muhammad Arifur Rahman - 6 years, 3 months ago

Congratulations for short sweet solution. But I think, a few only can understand the last step. The motion is equivalent to average vertical velocity with no gravity. You may explain it in a note below the solution..

Niranjan Khanderia - 6 years, 3 months ago

The problem is based on the fact the gravity will be acting on the bullet from the instant it leaves the gun barrel. The acceleration of the bullet will be $9.8\frac{m}{sec ^2}$ towards the earth for the entire 4.405 seconds of the shot. Without gravity the bullet would have traveled in a straight line hitting the wall at the same location as the aim point. The bullet accelerates vertically at the same rate do to gravity independent of the fact it is moving or not. Any object that is not self propelled and has no aerodynamic lift will be subject to gravitational acceleration the instant it is dropped, thrown or fired in any direction.

Darryl Dennis - 6 years, 3 months ago

Yes, I agree with you. But constant accelerated motion CAN be replaced by constant average velocity, and with velocity 0 at start and at mid flight, AT THESE TWO POINTS average velocity will be half the initial which is 1/2g(T\2) for both rise and fall periods. That is how your expression can be interpreted. T is full time. Or even as half the vertical height with constant velocity V*Sin $\theta$ if the time is for full rise and fall to ground. I hope I have made the point clear.

Niranjan Khanderia - 6 years, 3 months ago

It would have been a level 4 question if you hadn't mentioned that the mean velocity is the horizontal component ;)

Rohit Ner - 6 years, 3 months ago

I actually did consider posing the problem giving initial muzzle velocity and a crude air drag expression but decided not to. It seams very few people even attempt most of my problems as they are now. Making them even more difficult might result in no attempts at all.

Darryl Dennis - 6 years, 3 months ago

Well some thing has to be mentioned. Yes it could have been muzzle velocity.

Niranjan Khanderia - 6 years, 3 months ago

A long shot

The answer is 95.07.

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