Having studied projectile motion in physics lab, you decide to explore how the l

Question

Having studied projectile motion in physics lab, you decide to explore how the launch speed V0 (the speed of water at the tip of the water hose) and launch angle 0 would affect how long the water can stay in the air (air time) and how far the water can reach (range, R). Because the farm's surface is tilted by an angle from the horizontal, you define and measure the range R along the farm's tilted surface as shown in the figure above.

You will use the following 4-step strategy to solve this problem:
Step 1: Determine the X and Y coordinates of the water stream as a function of time, t.
Step 2: Determine the function that describes the sloped hill surface as a function of X and Y.
Step 3: Use your equations from steps 1 and 2 to solve for the air time, t, as a function of g, V0, 0, and .
Step 4: Use your calculated air time to calculate the range as R = (X² + Y²)½.

Step 1:
Consider the horizontal and vertical motion of the water stream independently. What is a symbolic expression for the X and Y coordinates of the water stream as a function of t, V0, 0, and the acceleration due to gravity, g?

STEP 2:

Next, consider the sloped surface of the hill. Write a symbolic expression for the equation that describes this slope. Answer in terms of X and .

Step 3:
You can now determine the intersection between the stream of water and the hill. Find when (in time) this intersection occurs by plugging your expressions from step 1 into your expression from step 2 to determine a symbolic expression for the air time, t, as a function of g, V0, 0, and .

Looking at your expression for air time, for a given slope and initial velocity, at what launch angle (limited to the "uphill" range of 0 to 90 degrees) would this time be maximized?
0=

Step 4:
Finally, use the expressions you have derived to this point to determine the range of the water stream, R = (X² + Y²)½. Answer symbolically in terms of g, V0, 0, and .
Also provide a numerical answer for the specific case of V0=13.4 m/s, 0=35.7 degrees, and =12.5 degrees. Use g=9.8 m/s2.

Explanation / Answer

step 1:

acceleration along X axis is 0 and acceleration along Y axis is -g.

formula used: distance=initial speed*time+0.5*acceleration*time^2

X=V0*cos(theta_0)*t

Y=V0*sin(theta_0)*t-0.5*g*t^2

step 2:

sloped surface is given by tan(omega)=Y/X

=>Y=X*tan(omega)

step 3:

using step 1 and step 2:

V0*sin(theta_0)-0.5*g*t^2=V0*cos(theta_0)*t*tan(omega)

==>0.5*g*t^2+V0*cos(theta_0)*tan(omega)*t-V0*sin(theta_0)=0

==>a*t^2+b*t+c=0

where a=0.5*g

b=V0*cos(theta_0)*tan(omega)

c=-V0*sin(theta_0)

this is a quadratic equation with one variable t and can be solved for t with given values.

t=(-b+sqrt(b^2-4*a*c))/(2*a)

here a is constant while b and c are variable.

t=(-V0*cos(theta_0)*tan(omega)+V0*sqrt(cos^2(theta_0)*tan^2(omega)+2*g*sin^2(theta_0))/g

t will be maximum when cos(theta_0)=0==>theta_0=90 degrees

step 4:

R=sqrt(X^2+Y^2)=sqrt((V0*cos(theta_0)*t)^2+(V0*sin(theta_0)*t-0.5*g*t^2)^2)

with the given values, t=3.607 seconds

then R=52.955 m

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