Aclam dropped by a seagull takes 3.0 seconds to hit the ground. what is the seagull's approximate height above the ground at the time the clam was dropped?

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

Put the value into the formula

We need to calculate the time the stick take to move

Using formula of time

Put the value into the formula

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.