Need help with this question regarding kinematics.

[DracoMalfoy]

Homework Statement

a) A ball is thrown in the -y direction off of a cliff with a velocity of 7m/s. If the ball takes 1.45s to reach the ground, how high off of the ground is the cliff? (Answer: -20m)

Homework Equations

Vf=Vi+a(t)

Displacement=Vi(t)+1/2a(t)^2

Vf^2=Vi^2+2a(Displacement)

Displacement= 1/2(Vf+Vi)(t)[/B]

The Attempt at a Solution

Known Values: [/B]

• Acceleration(a): -9.8m/s^2
• Time(t):1.45s
• Final Velocity(Vf): -7m/s
• Initial Velocity(Vi): 0m/s
• Displacement:?

I plug in the numbers and I'm not getting -20m as my answer. What am I doing wrong? or what am I not doing?

 

[Orodruin]

Initial Velocity(Vi): 0m/s

This is not correct. The problem statement says that the ball is thrown with an velocity 7 m/s in the negative y-direction. This refers to the initial velocity, not the final velocity.

 

[Charles Link]

The only equation you need is $s=v_i t +\frac{1}{2}at^2$. I don't know where the arithmetic is going wrong. Please show us what you get when you plug it in.

 

[DracoMalfoy]

This is not correct. The problem statement says that the ball is thrown with an velocity 7 m/s in the negative y-direction. This refers to the initial velocity, not the final velocity.

Yes, I see that now. I also didn't plug in the numbers correctly in my first attempt. I did end up getting -20 this time. If that really IS the correct answer. Thanks.

 

[CWatters]

Homework Statement

a) A ball is thrown in the -y direction off of a cliff with a velocity of 7m/s. If the ball takes 1.45s to reach the ground, how high off of the ground is the cliff? (Answer: -20m)

Homework Equations

Vf=Vi+a(t)

Displacement=Vi(t)+1/2a(t)^2

Vf^2=Vi^2+2a(Displacement)

Displacement= 1/2(Vf+Vi)(t)[/B]

Last line is wrong you lost the squares.

 

[Charles Link]

Last line is wrong you lost the squares.

Displacement is the average velocity multiplied by the time. The last line of the OP is correct.

 

[Orodruin]

Last line is wrong you lost the squares.

No, it is not wrong. Squares would make it dimensionally inconsistent.
Note that
$$
v_f^2 = v_i^2 + 2as \quad \Longrightarrow \quad
s = \frac{v_f^2 - v_i^2}{2a} = \frac{v_f+v_i}{2} \underbrace{\frac{v_f - v_i}{a}}_{= t}
$$