Question

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Answer 1

• Answer:
• Bullet misses the target
• Closest dist of approach ≈ 25 m
• Time of closest approach = 0.46 s
• Height of disc = 12.75 m

Explanation:

For disc A

u₁ = 50 m/s         θ₁ = 37°

∴ (u₁)$_{x}$ = 50 cos37° = 50 x 4/5 = 40 m/s

  (u₁)$_{y}$ = 50 x sin37° = 50 x 3/5 = 30 m/s

u$_{A}$ = 40 i + 30 j

For bullet B:

u₂ = 60√2 m/s        θ₂ = 45°

∴ (u₂)$_{x}$ = 60√2 cos45° = 60 m/s

  (u₂)$_{y}$ = 60 m/s

u$_{B}$ = 60 i + 60 j

Since the vertical components of u$_{A}$ and u$_{B}$ are not same the bullet does not hits the disc.

To find closest distance of approach :

Relative velo of B wrt A is

u$_{BA}$ = u$_{B}$ - u$_{A}$ = 20 i + 30 j

|u$_{BA}$| = $\sqrt{(400 + 900)}$ = 10$\sqrt{13}$  m/s

Angle made by u$_{BA}$ with horizontal  θ = tan⁻(3/2)

sinθ = 3/√13      and cosθ = 2/√13

Now Relative accln of B wrt A is

a$_{BA}$ = a$_{A}$ - a$_{B}$ = -g - (-g) = 0

∴ B moves with uniform velo along st. line wrt A ( see fig)

From fig

AB = initial separation between disc and gun = 30 m

The perpendicular dist from A to line of path of u$_{BA}$ is the closest

Closest dist of approach.

Closest dist of approach AC = AB sinθ

                                               = 30 x 3/√13

                                               = 90/√13

                                               = 25 m

BC = AB cosθ = 30 x 2/√13 = 60/√13 m

Time of closest approach t = BC/|u$_{BA}$|

                                             = (60/√13) / 10√13

                                             = 6/13

                                             = 0.46 s

Height of disc = (u₁)$_{y}$ t - 1/2 gt²

                       = 30x 0.46 - 5 x (0.46)²

                        = 13.8 - 1.05

                        = 12.75 m