Need help with this question regarding kinematics.

In summary: 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.In summary, the ball is thrown with an initial velocity of 0m/s and it takes 1.45s for it to reach the ground. The cliff is 20m off of the ground.
  • #1
DracoMalfoy
88
4

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?
 
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  • #2
DracoMalfoy said:
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.
 
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  • #3
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.
 
  • #4
Orodruin said:
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.
 
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Likes Charles Link
  • #5
DracoMalfoy said:

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.
 
  • #6
CWatters said:
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.
 
  • #7
CWatters said:
Last line is wrong you lost the squares.
No, it is not wrong. Squares would make it dimensionally inconsistent.
Note that
$$
v_f^2 = v_i^2 + 2as \quad \Longrightarrow \quad
s = \frac{v_f^2 - v_i^2}{2a} = \frac{v_f+v_i}{2} \underbrace{\frac{v_f - v_i}{a}}_{= t}
$$
 

1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion.

2. What are the three basic quantities in kinematics?

The three basic quantities in kinematics are position, velocity, and acceleration. Position describes the location of an object, velocity describes the rate of change of an object's position, and acceleration describes the rate of change of an object's velocity.

3. How is kinematics used in real life?

Kinematics is used in many real-life applications, such as designing roller coasters, analyzing the motion of vehicles, and predicting the trajectory of projectiles.

4. What is the difference between kinematics and dynamics?

Kinematics deals with the motion of objects without considering the forces that cause the motion, while dynamics involves studying the forces and their effects on the motion of objects.

5. How is kinematics related to calculus?

Kinematics involves the use of calculus to analyze the motion of objects, as calculus allows us to calculate instantaneous rates of change, such as velocity and acceleration, which are essential in kinematics.

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