Question

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.311 seconds to pass the length of the meter stick, how high ℎ0 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? Please define what your variables mean.

Answer 1

By using second kinematics,

S = u*t + 0.5*a*t^2

here, S = height of meterstick = 1.87 m

u = speed of acorn at top of the meter stick = ??

a = acceleration in vertical direction = g = 9.81 m/s^2

t = time taken to pass through length of the meter stick = 0.311 s

then,

1.87 = u*0.311 + 0.5*9.81*0.311^2

u = (1.87 - 0.5*9.81*0.311^2)/0.311

u = 4.49 m/s

Now, by using third kinematics law,

V^2 - U^2 = 2*a*h

here, here, V = final speed of acorn at top of meter stick = u = 4.49 m/s

U = initial pseed of acorn = 0 (As, free fall)

h = height of acorn above meter stick = ??

then,

h = (4.49^2 - 0^2)/(2*9.81)

h = 1.03 m

So,

Total height of acorn above the ground will be:

h0 = S + h

h0 = 1.87 + 1.03

h0 = 2.9 m

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