Basic kinematics - fox jumping over a wall problem

In summary, a fox is 3.2 m from a 0.5 m tall wall and leaps towards it with an initial velocity of 12.0 m/s at an angle of 18.9° above the horizontal. To find its height when it reaches the wall, the vertical and horizontal components of velocity must be calculated, as well as the time it takes for the fox to reach the wall. The correct equation to use is Δx = vxt and the missing term must be included. With the correct values for v_x and v_y, the time can be found and used to calculate the height using the equation Δy = 1/2at^2
  • #1
chococho
20
0
basic kinematics -- fox jumping over a wall problem

Homework Statement



A fox is 3.2 m from a 0.5 m tall wall. If it leaps from the ground towards the wall with an initial velocity of 12.0 m/s at angle of 18.9° above the horizontal, how high off the ground will the fox be when it reaches the wall?

Homework Equations



Δx = vxt
Δy = 1/2a2t

The Attempt at a Solution



First I tried to find Vy by doing "tan 18.9 = y/12"
and got 4.10 m/s.
Then I found the time by doing 3.2 = 12t
and got 0.266 sec.
I have no idea if I'm on the right track.
Then I just plugged in time and acceleration to find Δy and got 0.346 but it's wrong.
The answer is supposed 0.7 m...
Any help?
 
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  • #2
chococho said:

Homework Statement



A fox is 3.2 m from a 0.5 m tall wall. If it leaps from the ground towards the wall with an initial velocity of 12.0 m/s at angle of 18.9° above the horizontal, how high off the ground will the fox be when it reaches the wall?

Homework Equations



Δx = vxt
Δy = 1/2a2t

The Attempt at a Solution



First I tried to find Vy by doing "tan 18.9 = y/12"
and got 4.10 m/s.
Then I found the time by doing 3.2 = 12t
and got 0.266 sec.
I have no idea if I'm on the right track.
Then I just plugged in time and acceleration to find Δy and got 0.346 but it's wrong.
The answer is supposed 0.7 m...
Any help?
Welcome to PF chococho!

vy is the component of velocity in the vertical direction so it is just: v0sinθ

vx is the component of velocity in the horizontal direction so it is: v0cosθ

Using vx you can find t. Then use vy, factoring in the acceleration due to gravity, to determine how high the fox goes in the time t.

AM
 
  • #3
Second equation you used is missing a term.

Also tan does not give you vy.
 
  • #4
You have the initial velocity [itex] v = 12ms^{-1}[/itex] and the angle [itex] \theta = 18.9 °[/itex] with the horizontal,
find [tex]v_x = v \cos \theta [/tex] , its horizontal component, and [tex]v_y = v \sin \theta[/tex].Its vertical component. (That's where you were wrong.)
Then find time using the equation [tex] Δx =v_x t [/tex]
Then solve for [itex]Δy = \frac{1}{2} at^2[/itex]
 
  • #5
Thank you everyone. Kinematics is the first thing I learned in physics and I have a final coming up so I'm trying to re-learn this stuff... Still struggling :(
 
  • #6
chococho said:
Thank you everyone. Kinematics is the first thing I learned in physics and I have a final coming up so I'm trying to re-learn this stuff... Still struggling :(
Try to practice as many times as you can.
Kinematics is just like maths :smile:
 

What is basic kinematics?

Basic kinematics is the study of motion, specifically the position, velocity, and acceleration of objects without considering the forces that cause the motion.

What is the "fox jumping over a wall" problem?

The "fox jumping over a wall" problem is a classic example used to demonstrate basic kinematics. It involves a fox jumping over a wall with known height and distance, and the goal is to calculate the fox's initial velocity and time of flight.

What are the key variables in the "fox jumping over a wall" problem?

The key variables in this problem are the initial velocity of the fox, the height and distance of the wall, the acceleration due to gravity, and the time of flight.

How do you solve the "fox jumping over a wall" problem?

To solve this problem, you can use the kinematic equations of motion and plug in the known values to calculate the unknown variables. You can also use a kinematics calculator or graphing software to visualize the problem and find the solution.

What is the real-life application of the "fox jumping over a wall" problem?

The "fox jumping over a wall" problem can be applied to real-life situations, such as calculating the trajectory of a projectile, predicting the landing location of a jump, or analyzing the motion of athletes in sports like long jump or high jump.

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