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A stuntman jumps off the edge of a 45 meter tall building to an air mattress that has been placed on the street below at 15 meters from the edge of the building.


What minimum initial velocity does he need in order to make it onto the air mattress?

Sagot :

Eduard22sly

The minimum initial velocity needed by the stuntman in order to make it onto the air mattress is 5 m/s.

To obtain the answer to the question, we'll begin by calculating the time taken for the stuntman to get to the mattress. This can be obtained as follow:

Height (H) = 45 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

H = ½gt²

45 = ½ × 10 × t²

45 = 5 × t²

Divide both side by 5

[tex]t^{2} = \frac{45}{5}\\\\[/tex]

t² = 9

Take the square root of both side

[tex]t = \sqrt{9}[/tex]

t = 3 s

  • Finally, we shall determine the initial velocity needed by the stuntman. This can be obtained as follow:

Time (t) = 3 s

Distance of mattress from the building (s) = 15 m

Initial velocity (u) =?

s = ut

15 = u × 3

Divide both side by 3

[tex]u = \frac{15}{3}\\\\[/tex]

u = 5 m/s

Therefore, the minimum initial velocity needed by the stuntman in order to make it onto the air mattress is 5 m/s.

Learn more: https://brainly.com/question/24903556

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