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vladimir2022 [97]
3 years ago
6

If the vertical initial speed of the ball is 2.5 m/sm/s as the cannon moves horizontally at a speed of 0.55 m/sm/s , how far fro

m the launch point does the ball fall back into the cannon?
Physics
1 answer:
barxatty [35]3 years ago
7 0

Answer:

0.281 m

Explanation:

From vᵧ = uᵧ - gt

where vᵧ = final vertical component of the velocity

uᵧ = vertical component of the initial velocity = 2.5 m/s

g = 9.8 m/s²

At maximum height, vᵧ = 0 m/s

So,

Time to reach maximum height, t = uᵧ/g = 2.5/9.8 = 0.255 s

Total time of flight, T = 2 × time to reach maximum height

T = 2 × t = 2 × 0.255 = 0.510 s

Range of the cannon = uₓ T = 0.55 × 0.510 = 0.281 m

Note uₓ = horizontal component of the initial velocity.

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Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

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This means that the box moves 7.29cm backwards relative to the belt.

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