Answer:
The value is [tex]h_a = 1.712 \ m [/tex]
Explanation:
From the question we are told that
The height of the top meter stick above the ground is [tex]h_1 = 1.47 \ m[/tex]
The time taken for the acorn to pass the length of the stick is [tex]t = 0.281 \ s[/tex]
Generally the height of the acorn at the point it is the same height with the metered stick is mathematically represented as
[tex]h = h_m = u_a * t + \frac{1}{g} t^2[/tex]
Here [tex]h_m[/tex] is height of the meter stick and the value is 1 m (This because we are told in the question that the stick is 1 meter in length ( a meter stick))
So
[tex]1 = u_a * 0.281 + \frac{1}{9.8} (0.281)^2[/tex]
=> [tex]u_a = -2.2 \ m/s[/tex]
Generally the velocity of the acorn just before passing the top of the meter stick is mathematically represented by a kinematic equation as
[tex]u^2_a = u^2 + 2gs[/tex]
here u is zero since the acorn started from rest
So
[tex] (-2.2) = 0 + 2 * 9.8 * s[/tex]
[tex]s = 0.242 \ m[/tex]
Generally the height of the acorn is
[tex]h_a = h_1 + s[/tex]
[tex]h_a = 0.242 + 1.47[/tex]\
[tex]h_a = 1.712 \ m [/tex]
