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Anni [7]
3 years ago
14

A rock is launched vertically upward by a slingshot. The elastic band of the slingshot accelerates the rock from rest to 40 m/s

in a distance of 10 cm. How long will the rock take to return to its release point (time of flight)? Neglect air resistance.
Physics
1 answer:
eimsori [14]3 years ago
5 0

Answer:

0.016 s

Explanation:

Initial velocity, u = 0

use the third equation of motion to find the acceleration of the rock.

2as=v^2-u^2\\a = \frac{40^2-0}{2\times0.1} =8000 m/s^2

Net displacement in the vertical direction would be zero. y = 0

Use second equation:

s=ut+\frac{1}{2}at^2\\0=0+\frac{1}{2}(8000)t^2\\t= 0.016 s

Thus, the rock will return to its initial release point in 0.016 s.

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Rama09 [41]

Answer:

30.63 m

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So, substituting the values of the variables into the equation, we have

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