Question

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.07 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.121 seconds to pass the length of the meter stick, how high above the ground was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on the way down.

Answer 1

Answer:

5.05 m

Explanation:

length of stick (L) = 1 m

distance from the top of the stick to the ground (d) = 2.07 m

time taken for corn to travel the length of the stick (t) = 0.121 s

acceleration due to gravity (a) = 9.8 m/s^{2}

we can get the distance in four stages.

• Finding the velocity the corn was moving at from $s=vt+0.5at^{2}$

       where s = length of stick, t = time, a = acceleration due to gravity

       $s=vt+0.5at^{2}$

       1 = 0.121u + (0.5 x 9.8 x 0.121 x 0.121)

       1 = 0.121v +  0.0717

       1 - 0.0717 = 0.121v

       v = 7.67 m/s

• Using the velocity above to find the time it took for the corn to fall to the top of the stick from the formula v = u + at' where:

       v = final velocity = 7.67 m/s

       u = initial velocity = 0 m/s since it was initially at rest

       a = acceleration due to gravity

       t' = time taken to fall to the top of the stick

        7.67 = 0 + (9.8 x t')

        t' = 7.67 / 9.8 = 0.78 s

• using the time above to find the distance between the branch the corn falls from and the top of the stick using $s=vt+0.5at^{2}$ where:

        s = distance

        v = initial velocity = 0 m/s since it was initially at rest

        t = time = 0.78 s

        a = acceleration due to gravity = 9.8

        s = 0 + (0.5 x 9.8 x 0.78 x 0.78)

        s = 2.98 m

• Add the distance between the branch the corn falls from and the top of the stick and the distance from the top of the stick to the ground.

        2.98 + 2.07 = 5.05 m