Basic kinematics - fox jumping over a wall problem

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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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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
 
Second equation you used is missing a term.

Also tan does not give you vy.
 
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]
 
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 :(
 
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: