Question

In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the shell cracks. If a seagull flies to a height of 23.1 m, how long will the clam take to fall? The acceleration of gravity is 9.8 m/s 2 . Answer in units of sec.

Answer 1

Answer:

2.2 s

Explanation:

Hi!

Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

$x(t) = 23.1 m - \frac{1}{2}(9.8 m/s^2) t^2$

We are looking for a time t for which x(t) = 0

$0 = 23.1 m - (4.9 m/s^2) t^2$

Solving for t:

$t = \sqrt{\frac{23.1}{4.9}} s = 2.17124 s$

Rounding at the first decimal:

t = 2.2 s