Basic kinematics - fox jumping over a wall problem

[chococho]

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 = v_xt
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?

 

[Andrew Mason]

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 = v_xt
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!

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

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

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

AM

 

[paisiello2]

Second equation you used is missing a term.

Also tan does not give you vy.

 

[amind]

You have the initial velocity v = 12ms^{-1} and the angle \theta = 18.9 ° with the horizontal,
find v_x = v \cos \theta , its horizontal component, and v_y = v \sin \theta.Its vertical component. (That's where you were wrong.)
Then find time using the equation Δx =v_x t
Then solve for Δy = \frac{1}{2} at^2

 

[chococho]

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 :(

 

[adjacent]

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