Question

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

Answer 1

Answer:

Part a)

$\delta y = 0.85 m$

Part b)

$\theta = 0.65 degree$

Explanation:

Part a)

As we know that the target is at distance 75 m from the hunter position

so here we will have

$x = v_x t$

here we know that

$v_x = 180 m/s$

so we have

$75.0 = 180 (t)$

$t = 0.42 s$

now in the same time bullet will go vertically downwards by distance

$\delta y = \frac{1}{2}gt^2$

$\delta y = \frac{1}{2}(9.81)(0.42^2)$

$\delta y = 0.85 m$

Part b)

In order to hit the target at same level we need to shot at such angle that the range will be 75 m

so here we have

$R = \frac{v^2 sin(2\theta)}{g}$

$75 = \frac{180^2 sin(2\theta)}{9.81}$

$\theta = 0.65 degree$