+48 Problem # 8
projectile motion problems figure 2
A ball is kicked at an angle θ = 45°. It is intended that the ball lands in the back of a moving truck which has a trunk of length L = 2.5 m. If the initial horizontal distance from the back of the truck to the ball, at the instant of the kick, is do = 5 m, and the truck moves directly away from the ball at velocity V = 9 m/s (as shown), what is the maximum and minimum velocity vo so that the ball lands in the trunk. Assume that the initial height of the ball is equal to the height of the ball at the instant it begins to enter the trunk.
+118608 projectile motion problems figure 2
A ball is kicked at an angle θ = 45°. It is intended that the ball lands in the back of a moving truck which has a trunk of length L = 2.5 m. If the initial horizontal distance from the back of the truck to the ball, at the instant of the kick, is do = 5 m, and the truck moves directly away from the ball at velocity V = 9 m/s (as shown), what is the maximum and minimum velocity vo so that the ball lands in the trunk. Assume that the initial height of the ball is equal to the height of the ball at the instant it begins to enter the trunk.
You need to check my answer very carefully before you accept it. There is a lot of room for errors . 
That should have been 17.1 m/s and 18.8 m/s
+36915 OK....this is kinda long....hang in there and we'll see if I get the same answer as Melody.
The ball's x and y positions are given by
x=xo + vox t = voxt = .707 v t (1) (since xo=0 and 45 degrees and there is no x decceleration)
y= yo + voy t - 1/2 a t^2 = .707v t - 1/2 a t^2 (2)
The position of the two ends of the truck bed are given by
5m + 9 (t) and 7.5 + 9 (t)
Now...when the ball lands in the truck bed y will = 0, so
y=0= .707v t - 1/2 a t^2 (3)
and the 'x' values for the ball and truck bed will be equal
.707 v t = 5+ 9 t substitute THIS value of .707v t in to (3) and solve for t
0= 5 + 9t - 1/2 (9.8) t^2 results in t = 2.28 sec (using quadratic formula)
THEN the back of the truck bed will be at 5 + 9t = 25.55 m
(the front wil be 25.55+2.5=28.05m)
Now solve (1) for the ball velocity .707v(2.28)=25.55 results in v = 15.85 m/s
to land on the BACK of the truck bed.
Repeating some of this for the FRONT of the truck bed:
.707 v (2.28) = 28.05 = 17.4 m/s
Velocity must be 15.85-17.4 m/s (NOT QUITE THE SAME AS MELODY'S ANSWER)
Ooops..... GingerAle is correct....I constrained the motion of the truck for the second part (shown in THIS color above) ....here is the CORRECTED version for the second part (it is like the first part)
For the FRONT of the truck bed, we have .707 v t = 7.5 + 9t
Substitute THIS value in to equation (3) and solve for t
0= 7.5 + 9t - 1/2 (9.8)t^2 results in t= 2.459 sec (using quadratic formula)
Then the FRONT of the truck bed will be at 7.5 + 9t = 29.63 m
Now solve (1) for the ball velocity ,707 v (2.459) = 29.63 m results in v =17.04 m/s
to land on the FRONT of the truck bed
SOOOOO Velocity must be between 15.85-17.04 m/s Taa-Daa !
(Thanx everyone for pointing out the errors of my ways....hopefully I will not burn in hell forever)
+118608 Hi EP
It is good that we have done it different ways. One (or both) of us has probably made a little mistake as the answers should be closer. Still, the asker probably wanted one way or the other, now they have both. :)
+2440 No, you won’t burn in heII forever, just for 2.459 seconds in purgatory.
+2440 Hi Melody,
My jest was directed to EP, for his final (appended) remark, in the post above yours.
Purgatory is commonly regarded as a cleansing by way of painful temporal punishment, which, like the eternal punishment of h**l, is associated with the idea of fire. https://en.wikipedia.org/wiki/Purgatory#Pain_and_fire
Like any religious belief, individuals will adjust their viewpoints according to their own personality and perspective. So for many, such a place is an environment for prayer and meditation, while waiting for god to take them to heaven.
Personally, despite being raised as Catholic, I could never embrace any of the tenets of Catholicism (or any other religion). My first public demonstration of my contrary nature was in the fourth grade, after a visiting Brother gave a lecture (well couched in Catholic dogma) on creationism vs. evolution.
The teacher called on the students to ask questions or offer comments. Very few volunteered, so the teacher called on students individually. “Ginger, you usually have something to say.” I stood, which we were required to do when addressing the teacher or class, “You are too late,” I said. “I’ve already read Darwin’s book. God may have created the heavens and Earth, but I know I’m a chimp, just a new and improved version.”
After my comment, I started scurrying around the classroom and jumping on the tables while making chimp noises. I would have swung from the rafters if there were any. About a third of the class—mostly boys joined in the spectacle. I jumped on my best (boy) friend’s back; he could easily carry me around, and this continued for a few minutes, until the teacher scolded us back to our seats. During the commotion, I heard the Brother say to our teacher, “Your class room has become purgatory.” I remember thinking, “it may be purgatory for you, but it’s a little bit of heaven for us.”
Amazingly, we were not disciplined for this, although we did get a lecture on proper classroom behavior. After this, my classmates –usually boys– often brought me bananas. These banana gifts continued into high school. My chimp persona continues to this day.
Sister Alice, my mathematics teacher for 7th, 8th, and the first half of 9th grade, would often “condemn” her students (sometimes individually) to additional time in purgatory for poor math scores on tests. Her comments never came across as any kind of jest or levity. I would often intentionally “solve” problems incorrectly using various absurd methods, just to listen to her howl. My math grades were poor anyway, and by doing this, I rarely scored higher than D’s.
After one particular midterm exam, Sister Alice returned my test, saying, “Ginger, you are not going to purgatory, you’re going straight to heII!” To which I replied, “Oh, I’ll keep your company then, and you can teach me the correct way to solve these.”
My class thought this was funny, but Sister Alice didn’t. I received an immediate detention—the parochial school’s version of purgatory. I spent the time drawing Sister Alice roasting in heII, with demons stabbing her with pitchforks, inscribed with errant mathematical equations. I titled it Saint Alice Skips Purgatory.
Sister Alice was also the art teacher for the secondary grades, and she never cared much for my art, despite the fact that at least one of my art works always received honorable mention in every regional completion I entered—including one titled Beyond Purgatory, based on the one I did in detention. The Mother Superior (principal) confiscated the original—not because of the subject matter, but because detention was supposed to be a time of repentance and meditation. When she confiscated my drawing, she was doing her best to suppress her laughter.
The Mother Superior was usually “no nonsense” when it came to running the school, but she did have a sense of humor. Years later, after her funeral mass, someone told me she had my drawing framed, and she would hang it in her quarters for the week preceding All Saints' Day (Halloween).
To me this honor far exceeded all the gold/blue, silver/red, and bronze/white ribbons I’d won for any art completion.
GA
+112 Ginger, I will never tire of reading your narratives. Whether it’s a narrative from this world, or ... here.
+33614 Both methods are correct.
Melody has made a small numerical error in solving the last quadratic for reaching the rear of the truck. The value under the square root sign should be 601 rather than 761.
The other differences are due to small differences used for the numerical approximations used for 1/root2. (I get values for velocity between 15.83 and 17.04 m/s).
+2440 This is an interesting question.
Here’s a composite graph of the parabolas x=time for y=0 for the back and front of the tuck’s bed, and the corresponding velocities for these range times.
\(y=5+9t-\dfrac{1}{2}\cdot (9.8)t^2\\ y=7.5+9t-\dfrac{1}{2}\cdot (9.8)t^2\\ y=\text {velocity for time (t) to range target}\\ y=\frac{\left(t\cdot9.81\right)}{2\cdot\sin\left(45\right)}\\\)
---------
In EP's solution for the front of the truck bed, he adjusted the velocity by constraining the time component. In this case, with the increase in velocity the ball will pass over the front of the truck bed at time = 2.284, but it does not intersect y=0, here. In range equations, both time and velocity are functions of distance and changing one affects the other.
The ball with a velocity of 17.4m/s will cross y=0 in (17.4*2*sin(45)/9.81)= 2.51s at (17.4^2/9.8)= 30.86 meters. The target point of the truck will be at 7.5+(9*2.51)= 30.08 meters. So, the ball will land on the hood or the windshield.
GA
