Question

Basidiomycete fungi ballistically eject millions of spores into the air by releasing the surface tension energy of a water droplet condensing on the spore. The spores are ejected with typical speeds of 1.11 m/s, allowing them to clear the "boundary layer" of still air near the ground to be carried away and dispersed by winds, (a) If a given spore is accelerated from rest to 1.11 m/s in 7.40 us, what is the magnitude of the constant acceleration of the spore (in m/s) while being ejected? m/s(b) Find the maximum height of the spore (in cm) if it is ejected vertically. Ignore air resistance and assume that the spore is ejected at ground level. cm(c) Find the maximum horizontal range of the spore (in cm) if it is ejected at an angle to the ground. Ignore air resistance and assume that the spore is ejected at ground level CM

Answer 1

a) $1.5\cdot 10^5 m/s^2$

b) 6.3 cm

c) 12.6 cm

Explanation:

a)

The acceleration of an object is the rate of change of its velocity; it is given by:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time interval taken for the velocity to change from u to t

In this problem for the spore, we have:

u = 0 (the spore starts from rest)

v = 1.11 m/s (final velocity of the spore)

$t=7.40\mu s = 7.40\cdot 10^{-6}s$ (time interval in which the spore accelerates from zero to 1.11 m/s)

Substituting, we find the acceleration:

$a=\frac{1.11-0}{7.40\cdot 10^{-6}}=1.5\cdot 10^5 m/s^2$

b)

Since the upward motion of the spore is a free fall motion (it is subjected to the force of gravity only), it is a uniformly accelerated motion (=constant acceleration, equal to the acceleration due to gravity: $g=9.8 m/s^2$). Therefore, we can apply the following suvat equation:

$v^2-u^2=2as$

where:

v = 0 is the final velocity of the spore (when it reaches the maximum height, its velocity is zero)

u = 1.11 m/s is the initial velocity (the velocity at which it is ejected)

$a=-g=-9.8 m/s^2$ is the acceleration (negative because it is downward)

s is the vertical displacement of the spore, which corresponds to the maximum height reached by the spore

Solving for s, we find:

$s=\frac{v^2-u^2}{2a}=\frac{0^2-(1.11)^2}{2(-9.8)}=0.063 m = 6.3 cm$

c)

If the spore is ejected at a certain angle $\theta$ from the ground, then its motion is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion, with constant horizontal velocity

- A uniformly accelerated motion along the vertical direction (free fall motion)

The horizontal range of a projectile, which can be derived from the equations of motion, is given by:

$d=\frac{v^2 sin(2\theta)}{g}$

where

v is the initial velocity

$\theta$ is the angle or projection

g is the acceleration of gravity

From the equation, we observe that the maximum range is achevied when

$\theta=45^{\circ}$

For this angle, the range is

$d=\frac{v^2}{g}$

For the spore in this problem, the initial velocity is

v = 1.11 m/s

Therefore, the maximum range is

$d=\frac{(1.11)^2}{9.8}=0.126 m = 12.6 cm$