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The Cattle and Hay Problem

Let’s discuss this beautiful Cattle and Hay Problem useful for Physics Olympiad.

The Problem: The Cattle and Hay Problem

An airplane is dropping bales of hay to cattle. The pilot releases the bales at 150m above the level ground when the plane if flying at 75m/s in a direction above $55^\circ$ above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle is stranded?

Discussion:

The vertical part of the motion is projectile motion but the horizontal part is not.
The hay falls 150m with a downward acceleration g. It must travel a horizontal distance with constant horizontal velocity.
The bale has inital velocity components $$v_x=v_0cos\alpha=75cos55^\circ=43m/s$$
The vertical component is
$$v_y=v_0sin\alpha=75sin55^\circ=61.4m/s$$
Now, $y_0=150m$
For vertical motion we use the equation $$y-y_0=v_yt+\frac{1}{2}a_yt^2$$ Putting the appropriate values of $y$, $v_y$ and $a_y$, we have$$-150=6.14t-4.9t^2$$
. On solving the above quadratic formula for t, we get $$t=6.27\pm8.36$$.
Since t cannot be negative we take the positive value which is $t=14.6s$
Then putting the value of t in the equation for horizontal motion we have
$$x=v_xt+\frac{1}{2}a_xt^2$$
Now, $a_x=0$, so the second term in the equation becomes zero.
Hence $$x=(43)(14.6)=630m$$
So, the airplane needs to release the bales $630m$ in front of the cattle.

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The Snowball Problem

Let’s discuss the beautiful snowball problem useful for Physics Olympiad.

The Snowball Problem:

(a) A snowball rolls off a barn roof that slopes downward at an angle of $40^\circ$. The edge of the roof is 14.0m above the ground and the snowball has a speed of 7.00m/s as it rolls off the roof. How far from the edge of the barn does the snowball strike the ground if it doesn’t strike anything else while falling?

(b) A man 1.9m tall is standing 4.0m from the edge of the barn. Will he be hit by the snowball?

Solution:

The snowball follows the projectile motion.

In part(a), the vertical motion determines the time in the air.
The acceleration $$a_x=0$$, $$a_y=+9.80m/s^2$$ and $$v_x=v_0cos\theta_0=7cos40^\circ=5.36m/s$$
Vertical component of velocity $$v_y=v_0sin\theta_0=4.50m/s$$
Now, distance from ground $s=14m$.
$$14=(4.50)t+(\frac{1}{2}\times9.8\times t^2)$$
We use Sreedhar-Acharya’s method to solve for t.
$$t=\frac{(-4.5 \pm \sqrt{4.5^2-4(4.9)(-14)}}{2(4.9))}$$
The positive root for $t=1.29s$
The horizontal distance$$v_xt+(1/2)a_x t^2= 6.91m$$

(b) $$x-x_0=v_xt+\frac{1}{2}a_xt^2$$ gives $$t=\frac{x-x_0}{v_x}=\frac{4.0}{5.36}=0.746s$$ In this time the snowball travels downward a distance $$y-y_0=v_y+\frac{1}{2}a_y t^2=6.08m$$ and is therefore
$$14.0-6.08=7.9m$$ above the ground. The snowball passes well above the man and does not hit him.

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Suitcase Falling from an Airplane

Let’s discuss a beautiful problem useful for Physics Olympiad based on Suitcase Falling from an Airplane.

The Problem: Suitcase Falling from an Airplane

An airplane is flying with a velocity of $90m/s$ at an angle of $23^\circ$ above the horizontal. When the plane is $114m$ directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? Ignore air resistance.

Solution:

The suitcase moves in projectile motion. The initial velocity of the suitcase is equal to the velocity of the airplane.
To find the time, it takes to reach the ground
the y-component of the velocity $$v_{0y}=v_0sin23^\circ$$
The acceleration $$a_y=-9.8m/s^2$$ since +y is taken to be upward.
Now, the vertical distance from the plane to the dog s=114m.
Putting these values in the equation of motion to find the time t, we have
$$114=90sin23^\circ+\frac{1}{2}\times(-9.8)\times t^2$$
This gives, $$t=9.60s$$
TThe distance the suitcase travels horizontally is $$v_{0x}t=(v_0 cos23^\circ)t=795m$$

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Projectile Inside a Liquid

Let’s discuss a problem useful for Physics Olympiad, based on Projectile Inside a Liquid.

The Problem: Projectile Inside a Liquid

A body of mass m is projected inside a liquid at an angle θ0 with the horizontal at an initial velocity v0. If the liquid develops a velocity-dependent force F= -kv where k is a positive constant, determine the x and y components of the velocity at any instant.

Solution:

A body of mass m is projected inside a liquid at an angle θ0 with the horizontal at an initial velocity v0. The liquid develops a velocity-dependent force F= -kv where k is a positive constant.

Hence,

m dv/dt= -kv

or, dv/dt= -k/m v

or, dv/v= -k/m dt

Integrating both sides,

∫dv/v = -k/m ∫dt

ln|v|= -kt/m+c (where c is a constant of integration)….. (i)

Now, for the x component of velocity,

ln vx=  -kt/m+ c

From the given problem, we have

vx=v0 cosθ0 at t=0

Applying the above condition in eqn.(i), we get

ln (vx/ v0 cosθ0 )= -kt/m

or vx=v0cosθ0e-kt/m

For the y component, we have to consider the acceleration due to gravity g.

Hence,

m dvy/dt= -kvy-mg

or, dvy/(kvy+mg)= -k/m dt

Integrating both sides,

ln|kvy+mg|=-kt/m+c

At t=0, vy=v0sinθ0

Hence,

kvy+mg=(kv0sinθ0+mg) e-kt