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

Part a)

`\delta y = 0.85 m`

Part b)

`\theta = 0.65 degree`

Step-by-step 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 = (1)/(2)gt^2`

`\delta y = (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 = (v^2 sin(2\theta))/(g)`

`75 = (180^2 sin(2\theta))/(9.81)`

`\theta = 0.65 degree`