Acanister is released from a helicopter 500m 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.

a) h = 1/2 gt² b) 99 m/s c) 14.1 m/s

Explanation:

a) As the only force acting on the canister once released is gravity, as it is constant, we can use one of the kinematic equations, as follows:

h = h₀ + v₀t + 1/2a t²

As we are told that the canister is simply released, it means that v₀ =0.

If we choose the direction of the acceleration (downward) as positive ,and we select the height at which it was released as our origin, so h₀ =0, the final expression for height is as follows:

h = 1/2 g t²

b) Combining this equation with the expression of t, from the definition of acceleration as the rate of change of velocity, we arrive to this expression:

vf² - v₀² = 2 g h  

As v₀ = 0, we have v₀ = √2.g.h = √2.9.8.500 m²/s² = 99 m/s

c) In order to be able to break the canister, impact speed must be larger than 100 m/s.

So, we can use the same equation as above, putting vf=100 m/s, and solving for v₀, as follows:

v₀² = vf² - 2.g.h = 100² m²/s² - 2.9.8.500 m²/s² = 200 m²/s²

v₀ = √200 m²/s² = 14.1 m/s

For any value of v₀ just barely larger than this, the canister will be broken.