Question

You drop a seashell into the ocean from a height of 40 feet. Write an equation that models the height h in feet) of the seashell above the water after seconds. How long is the seashell in the air? Round your answer to the nearest tenth.

Answer 1

Answer:

t=1.58 seconds

Explanation:

The equation which models the height h of the seashell above water after t seconds

Answer 2

Height of seashell above water after t seconds; h = 40 - 8t²

Time that seashell is in air; t = 1.58 s

The sea shell is dropped under gravity and thus, from Newton's equations of motion, we know that height/distance is;

S = ut + ½gt²

Since it was dropped from a height and not the ground, then;

h = height of shell from ground

h_o = height from which the shell was dropped.

Thus, we have;

h - h_o = ut + ½gt²

We are told that the seashell was dropped into the ocean from a height of 40 ft. Thus;

h_o = 40 ft

It was dropped from rest and so; u = 0 ft/s

Acceleration due to gravity is; g = -32ft/s² (since it is free fall)

Thus;

h - 40 = 0(t) + ½(-32t²)

h = 40 - 8t²

The time that the seashell in the air will be after it has hit the surface which will be when h = 0.

Thus;

0 = 40 - 16t²

16t² = 40

t² = 40/16

t² = 2.5

t = √2.5

t = 1.58 s

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