Answer to Question #114676 in Quantum Mechanics for Michael Mateus

Question

5. A hunter aims directly at a target (on the same level) 75.0 m.
a) If the bullet leaves the gun at speed of 180m/s by how much will it miss the target?
b) At what angle should the gun be aimed so as to hit the target?

Answer 1

If the bullet is fired horizontally in the same level as that of the target it will be pulled by the gravitational force during the flight.

Initial velocity = 180 m/s

Distance of the target = 75 m

Time taken = dist. / vel.= 75/180 = 0.416 sec

During this time the bullet is under the gravitational pull

Its vertical initial velocity = u = 0 m/s

Acceleration = a = 9.8 m/s²

Time = 0.416 sec

The distance covered $h = ut + \frac12at²$

h = $0 \times 0.416 + \frac12 (9.8) (0.416)^2$

h = 0 + 0.85 m

The bullet will miss the target by 0.85 m

To adjust this the bullet should be aimed at a little elevated, say θ°

It will be a projectile motion. During the projectile motion the "time of flight" is equal to

T =$\dfrac{(2 u Sin θ) }{g,}$

& the "horizontal range" R is equal to:

R = $\dfrac{(u² Sin 2θ) }{ g}$

where u is the initial velocity, θ, the angle of elevation., g, the acceleration due to gravity.

75 = $\frac{(180^2\sin 2θ) }{9.8}$

75 = $\frac{32400 \times \sin 2θ }{ 9.8}$

Sin 2θ =$\frac{ (75 \times9.8) }{ 32400}$ = 0.0227

2θ = 1.3°

θ = 0.65 °

The barrel should be elevated through an angle 0.65°