Question

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn falls from somewhere higher up in the tree. if the acorn takes 0.301 seconds to pass the length of the meter stick, how high above the ground was the acorn before it fell (assuming that the acorn didn\'t run into any branches or leaves on the way down)?

Answer 1

The acorn was at a height of 4.15 m from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

Answer 2

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.301 seconds to pass the length of the meter stick, the distance showing how high above the ground was the acorn before it fell 2.449 m

From the given information,

• the height of the meter stick is = 2.27 meters above the ground
• the time it takes the acorn to pass the length of the meter stick after falling = 0.301 seconds

From here, we know our time (t)  = 0.301 seconds, the distance (s) = 1 m

Now, use the second equation of motion to determine our final velocity.

$\mathbf{S = ut+ \dfrac{1}{2}at^2}$

$\mathbf{(1) = u(0.301)+ \dfrac{1}{2}(9.8 )(0.301)^2}$

$\mathbf{(1) = u(0.301)+0.4439}$

1 - 0.4439 = 0.301 u

0.5561 = 0.301 u

u = 0.5561/0.301

u = 1.847 m/s

Now, the height of the acorn above the meter stick can be estimated as:

v² = u² + 2gh

Making height the subject of the formula:

$\mathbf{h = \dfrac{v^2 - u^2}{2g}}$

$\mathbf{h = \dfrac{1.837^2 - 0^2}{2(9.8)}}$    since the initial velocity (u) is 0

h = 0.179 m

Therefore, we can conclude that the height of the acorn from the top of the tree past the meter stick till it reaches the ground is (0.179 m + 2.27 m) = 2.449 m

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