Acanister is dropped from a helicopter 500m above the ground. its parachute does not open, but the canister has been designed to withstand an impact velocity of 100 m/s. will it burst?

The impact speed 98.995 m/s is less than 100 m/s and the canister will not burst.

Step-by-step explanation:

A function F is called an antiderivative of f on an interval I if for all x in I.

Recall that if the object has position function , then the velocity function is . This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration function is , so the velocity function is an antiderivative of the acceleration.

An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by . For motion close to the ground we may assume that is constant, its value being about .

We know that the acceleration due to gravity is given by

and the antiderivative is velocity

We know that the canister was dropped, so the initial velocity at t = 0 is zero, this fact let us know the value of C.

The antiderivative of velocity is the position

To find the value of the constant C, we know that the height was 500 m at t  = 0, this means

Using the fact that at the time of impact the height s(t) is zero we can compute the total time of the fall:

A negative time does not make sense, so we only take as a possible solution

Now the final velocity is

The impact speed 98.995 m/s is less than 100 m/s and the canister will not burst.

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.