Projectile motion snowball problem

[guitarguybri]

Homework Statement

One child is trying to hide from another who is throwing snowballs. The first child hides by crouching behind a 2.5 m high fence that is 4 m from the other child's snow fort. The second child tries to hit the hiding child by lobbing snowballs over the fence as shown in the figure. If the maximum speed with which the second child can throw a snowball is 10 m/s, what is the closest distance to the backside of the fence that he can get a snowball to land? (Assume the snowball is released at a height equal to that of the back of the fence.)

http://mygateway.umsl.edu/courses/1/UMSL-BUSAD3320-005-41676-200743/uploads/homepage/_1778814_1/problem.jpg

vi = 10m/s
delta_y= 2.5m
delta_x = x + 4m
g = 9.8 m/s2
yf = 0 m

t - ?
xf - ?

Homework Equations

yf = vyi*t - .5*g*t^2 --> vi*(sin theta)*t - .5*g*t^2
xf = vxi*t --> vi*(cos theta)*t

t = xf / (vi*cos theta)

after much algebra and trig substitutions (I got rid of t in the yf equation).

0 = ((g*(xf)^2)*(tan theta)^2)/(2*(vi)^2) -xf*tan theta + (g*(xf)^2)/(2*(vi)^2) + yf

(-b +/- sqrt((b)^2 - 4ac)) / 2a -- quadratic formula

The Attempt at a Solution

I want to use the second to last equation I listed. I want to use the quadratic formula to solve for tan theta. Then I can take the inverse tangent and get the angle. Then I can use kinematic equations to find the distance that I want. Which will be 4m + x. But I have too many unkowns and not enough equations. I want to know how to find out what xf is. I think I can handle it after that. I just need to be able to use the quadratic formula to find my angle theta. Thanks!

 

[anathema]

I would first gather all the necessary information from the given scenario: the height of the fence, the distance between the two children, and the maximum throwing speed of the second child. I would also assume that the snowball is thrown at an angle that is parallel to the ground, since it is being lobbed over the fence.

Next, I would set up a coordinate system with the origin at the base of the fence and the x-axis along the ground towards the hiding child. This will help me visualize the problem and keep track of my variables.

Using the kinematic equations, I can find the time it takes for the snowball to reach the top of the fence (since it is released at the same height as the back of the fence) and the distance it travels in the x-direction:

t = delta_y / (vi*sin theta)
xf = vi*cos theta * t

Now, to solve for theta, I would use the quadratic formula as you suggested. However, I would set the equation equal to 0 and solve for tan theta:

0 = (g*xf^2*tan^2 theta) / (2*vi^2) - xf*tan theta + (g*xf^2) / (2*vi^2)

This equation can be simplified to a quadratic equation in terms of tan theta:

0 = (g*xf^2) / (2*vi^2) * (tan^2 theta - 2*xf*tan theta + 1)

Using the quadratic formula, we can solve for tan theta:

tan theta = (xf +/- sqrt(xf^2 - (g*xf^2)/(vi^2))) / (g*xf)/(vi^2)

Since we are only interested in the positive solution, we can ignore the negative sign in front of the square root. Plugging in the given values, we can solve for tan theta:

tan theta = (4 +/- sqrt(4^2 - (9.8*4^2)/(10^2))) / (9.8*4)/(10^2)

tan theta = 0.55

Taking the inverse tangent, we get the angle theta to be approximately 30 degrees.

Now, we can use this angle to find the distance from the back of the fence where the snowball will land:

x = xf + 4m = (vi*cos theta * t) + 4m = (10

 

[ogb p]

I would recommend approaching this problem by breaking it down into smaller parts and using the relevant equations for each part. First, we can determine the time it takes for the snowball to reach the top of the fence using the equation yf = vyi*t - .5*g*t^2. We know that yf = 2.5m, vyi = 10m/s, and g = 9.8m/s^2. Solving for t, we get t = 0.71 seconds.

Next, we can use this time to determine the horizontal distance the snowball travels using the equation xf = vxi*t. We know that vxi = 10m/s and t = 0.71 seconds, so xf = 7.1m. This is the distance from the base of the fence to where the snowball reaches its maximum height.

To determine the angle at which the snowball is thrown, we can use the equation xf = (vi^2*sin2(theta))/g. We know that xf = 7.1m, vi = 10m/s, and g = 9.8m/s^2. Solving for theta, we get theta = 26.6 degrees.

Finally, we can use this angle to determine the distance from the backside of the fence that the snowball will land. Using the equation xf = vi*cos(theta)*t, we can solve for x and add it to the 4m distance from the backside of the fence to get the total distance.

In summary, to solve this problem we used the equations yf = vyi*t - .5*g*t^2, xf = vxi*t, and xf = (vi^2*sin2(theta))/g to determine the time, horizontal distance, and angle at which the snowball is thrown. Then, we used the equation xf = vi*cos(theta)*t to determine the distance from the backside of the fence that the snowball will land.