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A canister is released from a helicopter 500 m above the ground. The canister is designed to withstand an impact speed of up to 100 m/s.
(a) Ignoring air resistance, find an equation of the height at any time t.
(b) Find the impact speed of the canister.
(c) Your answer to (b) should be less than 100 m/s.
Rather than just releasing it, let’s see if we could break it by throwing it down with an initial velocity. Re-do parts (a) and (b) with an initial velocity v0, then find the value of v0 required to break the canister.

Respuesta :

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Answer:

(a) h = 500 - 4.9t²

(b) 98.99 m/s

(c) h = 500 - Vot - 4.9t²

    14.14 m/s

Explanation:

(a) You can use the equation of linear motion s = vi.t + 0.5gt²

(b) [tex]v_{f} ^{2} = v_{i} ^{2} +2ax\\v = \sqrt{0^{2} + 2*9.8*500 } = 98.99 m/s[/tex]

(c) [tex]v_{f} ^{2} = v_{i} ^{2} + 2ax\\v_{i} = \sqrt{100^{2} -2*9.8*500}=14.14 m/s[/tex]

The initial velocity mucg be greater than 14.14 m/s to break the cannister

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