Projectile problem-A rock is kicked off a 45 Degree hill at 15m/s.

In summary, the conversation is about a projectile problem where a rock is kicked off a 45 degree hill at 15m/s. The problem is to find how long it takes for the rock to hit the ground. Different methods were tried, including equating the X and Y position functions and using kinematic equations. However, the given answer in the back of the book did not match with the answer obtained. It was later discovered that there was a misprint in the problem and the actual velocity should be 25 m/s.
  • #1
Kizaru
45
0
Projectile problem--A rock is kicked off a 45 Degree hill at 15m/s.

Hello. I tried this problem and it's the last one for my problem set due Thursday, but my answer differs from the given answer in the back of the book. After trying the solution again, I searched Google and found this. I'm having the exact problem.

Homework Statement


A rock is kicked horizontally at 15m/s from a hill with a 45 degree slope. How long does it take for the rock to hit the ground?
Vix = 15m/s
Viy = 0
Yi = 0
Xi = 0

Homework Equations


Vx = Vix
X = Xi + Vxo*t
Vy = Viy - gt
Y = Yi + Viy*t - (1/2)g*t^2


The Attempt at a Solution


I tried equating the X and Y position functions, but this only provides the intersection between them--it does not provide the intersection between the rock and the hill. My second method is identical to the one described in the link mentioned earlier in my post. The problem is that my answer isn't the same as the one in the back of the book, and I'm not sure if it's a problem on my end. I don't know what to do after this. Any help would be great :)
 
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  • #2


If the rock lands at the same height it was launched one can obtain the time from the y position equation since y=0.
 
  • #3


Kurdt said:
If the rock lands at the same height it was launched one can obtain the time from the y position equation since y=0.

The rock will never pass that point because it's launched horizontally (no initial Y velocity) off a 45 degree hill (the hill is declining). The only time the rock is at y=0 is when t=0.

Imagine graphing y=-x and y=-x^2. If you focus on only the third quadrant, you have what looks like the hill (y=-x) and the path of the rock (y=-x^2), but it's not that easy to find an equation that represents the path of the rock.
 
  • #4


Oh ok I get it. I thought it was launched off a 45 degree slope.

Well you know how the x position and y position are related (equation of slope) and you know how they evolve over time (kinematic equations). You can use both these bits of info to find the time.
 
  • #5


For those getting an answer of 3.06 seconds, which disagrees with the back of the book, this is because there was a misprint. The actual velocity should be 25 m/s, not 15 m/s. The solutions manual also uses 25 m/s. Good luck!
 

1. What is the initial velocity of the rock?

The initial velocity of the rock is 15 m/s, as stated in the problem.

2. What is the angle at which the rock is kicked off the hill?

The rock is kicked off the hill at a 45 degree angle, as stated in the problem.

3. How can we calculate the distance traveled by the rock?

The distance traveled by the rock can be calculated using the formula: d = v2sin(2θ)/g, where v is the initial velocity, θ is the angle, and g is the gravitational acceleration (9.8 m/s2).

4. What other factors can affect the trajectory of the rock?

Other factors that can affect the trajectory of the rock include air resistance, wind, and the shape/weight of the rock itself.

5. Can we determine the final velocity of the rock when it reaches the ground?

Yes, we can use the formula: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time taken for the rock to reach the ground.

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